7 concepts23 worked examples28 exercises3 exam-level7 figures
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12Inverse trigonometric functions, hyperbolic functions, and indeterminate forms revisited
A student checks $e$ the slow way: take $1+\tfrac{1}{1000}$ and raise it to the power $1000$. Every extra factor moves the answer by a tenth of one percent, so it should stay near $1$. The calculator returns $2.7169$.
By the end you can name what kind of fight that expression is, turn it into a quotient a derivative can attack, and get the exact value.
In 60 seconds
Three new derivative families — general bases, inverse trigonometric, hyperbolic — come out of one idea, invert then differentiate; l'Hôpital's rule handles the limits they produce.
The constant is in the base and the variable in the exponent, or the logarithm is not natural. The inner derivative $u'$ is the chain rule and is there for every base; the natural base is the same rule with $\ln e=1$, so nothing is left behind but $u'$.
Substitution gives $0/0$ or $\infty/\infty$, and only then.
Three most common mistakes
Writing $(f^{-1})'(b)=1/f'(b)$. The denominator is $f'$ at the point that maps to $b$, not at $b$ itself.
Applying l'Hôpital's rule to a limit that is not indeterminate. The rule then hands back a wrong number instead of a warning.
Taking the logarithm of a power form and then reporting $\ln L$ as the answer. The last line of that method is always an exponential.
The weighting sheet gives two midterms at $28\%$ each, a final at $28\%$, quizzes $10\%$, homework $6\%$. It says nothing about how many questions come from any one week.
How much time do you have?
10 minutes
You leave able to differentiate the six standard inverses and to name the form of a limit before touching it.
In 60 seconds card, Reciprocal slopes: the derivative of an inverse, Naming the form and choosing the move, Formula card
45 minutes
Add the derivations, the sign family and the logarithm trick; you can now answer a full multi-part question instead of quoting a table.
everything in the 10 minute path, Restricting the sine until it can be inverted, Where the minus signs come from, l'Hôpital's rule and the two forms it accepts, The five forms that need a rewrite first, Scaffolding comes off, Full exam-style question
full reading
Everything above plus the general bases, the hyperbolic family, and practice graded from concept traps to problems that hide their type.
all blocks in order, Practice A, B, C and D, Mistake ledger, Check yourself
By the end of this section
Compute the derivative of an inverse function at a point from the derivative of the original function, without a formula for the inverse.
Differentiate $a^{x}$, $\log_a x$ and $x^{n}$ for a real exponent by rewriting each one through $e$ and $\ln$.
Derive the derivatives of $\arcsin$ and $\arctan$ by implicit differentiation on the , and apply them with the chain rule.
Explain where the minus signs and the absolute value in the inverse trigonometric derivative table come from, and state the interval on which each formula is valid.
Manipulate $\sinh$, $\cosh$ and $\tanh$ through their exponential definitions, and read an as a logarithm.
Decide whether l'Hôpital's rule may be applied to a given limit, and apply it repeatedly with a form check between applications.
Convert $0\cdot\infty$, $\infty-\infty$, $0^{0}$, $1^{\infty}$ and $\infty^{0}$ into a quotient the rule accepts, and undo the conversion at the end.
Syllabus coverage
6.4*
General logarithmic and exponential functions
Bases other than $e$: the derivatives of $a^{x}$ and $\log_a x$, the power rule for a real exponent, and the limit that produces $e$.
covered
6.8
and l'Hôpital's rule
The statement and its hypotheses, why it holds for $0/0$, the two ways it is misused, and the five forms that need a rewrite first.
covered
6.6
Principal branches, the derivations of the derivatives, the sign family from the complementary identity, and the interval each formula holds on.
covered
Hyperbolic and inverse hyperbolic functions
The week's syllabus line names no book section for these, so none of it is required material. It is here because the section title names them and because their inverses are the one place the reciprocal-slope argument can be checked against a closed form.
Every derivative in this section is a composite, and the inner factor is the one that gets dropped.
Implicit differentiation
If $y$ is a function of $x$, then $\dfrac{d}{dx}F(y)=F'(y)\cdot\dfrac{dy}{dx}$; differentiate both sides of an equation and solve for $\dfrac{dy}{dx}$.
It is the engine behind every inverse derivative here: we never solve for the inverse, we differentiate the equation that defines it.
Derivatives of the natural pair
$\dfrac{d}{dx}\ln x=\dfrac1x$ for $x>0$, and $\dfrac{d}{dx}e^{x}=e^{x}$.
Both were built in the previous section, and every general base is rewritten through them.
Pythagorean identities
$\sin^{2}\theta+\cos^{2}\theta=1$ and $1+\tan^{2}\theta=\sec^{2}\theta$.
They convert $\cos(\arcsin x)$ and $\sec^{2}(\arctan x)$ into algebra, which is where the roots in the table come from.
Monotone means invertible
If $f'>0$ on an interval, then $f$ is strictly increasing there, hence one to one and invertible on that interval.
Every inverse in this section exists only because a domain was cut down to a piece where the derivative keeps one sign.
Addition formula for the sine
$\sin(A+B)=\sin A\cos B+\cos A\sin B$; with $B=\tfrac{\pi}{4}$ it gives $\sin x+\cos x=\sqrt2\,\sin\!\left(x+\tfrac{\pi}{4}\right)$.
It opens a shifted angle in one line, which is the short route through a limit that would otherwise need three applications of l'Hopital's rule.
Try it yourself first (2 questions)
1§12.0 — implicit differentiation, one line of it●●○○○
Everything in this section is built by differentiating an equation in which $y$ is not isolated. This is that step, on its own.
Given
$\sin y=x$, where $y$ is a function of $x$
Find
(a) Which line is the result of differentiating both sides with respect to $x$?
Hint 1/4
Ask what $y$ is here: a number or a function of $x$? That decides whether differentiating $\sin y$ needs the chain rule.
Hint 2/4
The chain rule on the left: $\dfrac{d}{dx}\sin y=\cos y\cdot\dfrac{dy}{dx}$, and the right side differentiates to $1$.
Hint 3/4
So the equation becomes $\cos y\cdot y'=1$, with $y'$ the unknown to be isolated.
Hint 4/4
The correct line is $\cos y\cdot y'=1$.
Show solutionDifferentiate both sides
$$\frac{d}{dx}\sin y=\cos y\cdot\frac{dy}{dx}$$
$y$ depends on $x$, so the chain rule applies on the left
$$\frac{d}{dx}x=1$$
the right side
$$\cos y\cdot y'=1$$
set the two equal
Answer $$\boxed{\cos y\cdot y'=1}$$
Check
Check on a case you can solve: $y=\arcsin x$ at $x=0$ gives $\cos 0\cdot y'=1$, so $y'=1$ — and $\arcsin$ does leave the origin with slope $1$.
2§12.0 — the natural logarithm and the chain rule●●○○○
The general bases here are all built on the natural logarithm, so this derivative has to be automatic before the new material starts.
Given
$y=\ln\left(x^{2}+1\right)$
Find
(a) Compute $y'$ and evaluate it at $x=2$.
Hint 1/4
Identify the inner function; the logarithm is applied to something that is not just $x$.
Hint 2/4
$\dfrac{d}{dx}\ln u=\dfrac{u'}{u}$.
Hint 3/4
Here $u=x^{2}+1$ and $u'=2x$, so $y'=\dfrac{2x}{x^{2}+1}$; the point asked for is $x=2$.
Hint 4/4
$y'(2)=\dfrac{4}{5}$.
Show solutionChain rule
$$y'=\frac{2x}{x^{2}+1}$$
outer derivative $1/u$, inner derivative $2x$
$$y'(2)=\frac{4}{5}=0.8$$
substitute
Answer $$\boxed{y'(2)=0.8}$$
Check
Numerical check with a symmetric difference across $x=2$: $\ln\left(2.01^{2}+1\right)-\ln\left(1.99^{2}+1\right)=1.617426-1.601426=0.016000$, and the two points are $0.02$ apart, so the measured slope is $0.800$.
Notation
symbol
reads as
means
watch out
$\arcsin x$
arc sine of x
the angle in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ whose sine is $x$
also written $\sin^{-1}x$; the $-1$ means inverse function, never $1/\sin x$
$\arctan x$
arc tangent of x
the angle in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ whose tangent is $x$
defined for every real $x$, unlike the arcsine
$\log_a x$
log base a of x
the exponent to which $a$ must be raised to give $x$
to differentiate it, first rewrite it as $\ln x/\ln a$
$\sinh x,\ \cosh x$
hyperbolic sine and hyperbolic cosine of x
the odd and even halves of $e^{x}$
neither is periodic and $\cosh x\ge 1$ always; the resemblance to the circular pair is in the algebra, not the graphs
$\operatorname{sech} x$
sech of x
$1/\cosh x$
appears squared in $(\tanh x)'$, and $\operatorname{sech}^{2}x=1-\tanh^{2}x$
$0\cdot\infty,\ \infty-\infty,\ 1^{\infty}$
the indeterminate shapes
shorthand for how the two parts of a limit behave, not arithmetic on numbers
a shape is a question; writing $1^{\infty}=1$ answers it without doing the work
Conventions used here
Angles are radians
Every derivative formula here assumes radians. In degrees the sine has derivative $\tfrac{\pi}{180}\cos x$ and the whole table changes.
The formulas descend from $\lim_{h\to 0}\tfrac{\sin h}{h}=1$, which is true only in radians.
Which branch an inverse returns
$\arcsin$ returns a value in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, $\arccos$ in $[0,\pi]$, $\arctan$ in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. When a question asks for an angle, that is the one meant.
$\sin\theta=\tfrac12$ has infinitely many solutions; a function must pick one, and that choice is what makes the derivative formulas well defined.
The secant case is a convention, and we say so
For $x>1$ every book agrees that $\left(\operatorname{arcsec} x\right)'=\dfrac{1}{x\sqrt{x^{2}-1}}$. For $x<-1$ the sign depends on the branch the book chose, so here the secant inverse appears only where the question states its range.
An unstated branch turns a sign question into a coin toss, and a coin toss is not a solution.
How a limit is reported
Writing $\lim=+\infty$ says the limit does not exist and says how it fails. We keep it when true, and reserve does not exist for values that oscillate.
The infinite statement carries information that the bare denial throws away.
Indeterminate is about limits, not arithmetic
$0^{0}$, $1^{\infty}$, $\infty^{0}$, $0\cdot\infty$ and $\infty-\infty$ name the behaviour of a limit, not an arithmetic value; as limits they can come out to anything.
Two limits with the same shape were shown here to give $e$ and $e^{3}$, so the shape cannot be the answer.
Where a derivative formula stops
$(\arcsin x)'$ is quoted for $-1<x<1$ only and $\left(\cosh^{-1}x\right)'$ for $x>1$ only: endpoints with a vertical tangent are excluded even though the function is defined there.
Substituting an endpoint divides by zero, and the division by zero is the graph reporting a vertical tangent.
12.1Reciprocal slopes: the derivative of an inverse
Gets the derivative of an inverse at one point from $f'$ alone, when you cannot solve $y=f(x)$ for $x$.
The last section built one inverse pair by hand. This one needs six more, and none can be solved for with algebra, so the first job is to differentiate an inverse we cannot write down.
Invert $y=3x+5$ by hand and differentiate the result.
Differentiate $\ln x$, because the last section built it.
Not solvable yet
Solve $y=x^{3}+2x+1$ for $x$, so the inverse has no formula to differentiate.
Solve $x=\sin y$ for $y$ with algebra.
Write $\arcsin x$ with roots and quotients.
The graphs of $f$ and $f^{-1}$ are mirror images across $y=x$, so perhaps the slopes are mirror images too and $\left(f^{-1}\right)'(b)=\dfrac{1}{f'(b)}$.
Why it fails
Reflection does reciprocate the slope — it swaps rise and run — but it also moves the point, and the naive formula forgets that. With $f(x)=x^{3}+2x+1$ and $b=4$: since $f(1)=4$, the true value is $1/f'(1)=\tfrac{1}{5}$ and the naive $1/f'(4)=\tfrac{1}{50}$ is ten times too small.
TheoremTheorem 12.1: derivative of an inverse function
Conditions
$f$ is one to one and differentiable on an open interval $I$
The slope of the inverse at an output equals one over the slope of the original function at the input that produced that output.
Where it comes from
Differentiate $f\!\left(f^{-1}(x)\right)=x$ with the chain rule: $f'\!\left(f^{-1}(x)\right)\cdot\left(f^{-1}\right)'(x)=1$, then divide. What this does not prove is that $f^{-1}$ is differentiable at all — that is what the hypothesis $f'(a)\neq 0$ buys.
Reflection swaps the coordinates of a point and therefore swaps rise with run: a steep slope becomes a shallow one at the mirrored point, never at the same point.
Looks like this, but is not
$f(x)=x^{3}$ is one to one on the whole line, differentiable everywhere, and its inverse $x^{1/3}$ is written down explicitly. So the theorem should apply at $b=0$.
At $a=0$ we have $f'(0)=0$ and the conclusion fails: $x^{1/3}$ has a vertical tangent at the origin, so $\left(f^{-1}\right)'(0)$ does not exist. A zero slope reflects into a vertical one.
The slope of the inverse of x cubed plus 2x plus 1 at the output 4
$f(x)=x^{3}+2x+1$ has no usable inverse formula: solving $y=x^{3}+2x+1$ for $x$ needs the cubic formula. We want the slope of that inverse at $y=4$.
Given
$f(x)=x^{3}+2x+1$
the output $b=4$
Find
$\left(f^{-1}\right)'(4)$
SolutionCheck that an inverse exists
$$f'(x)=3x^{2}+2\ge 2>0$$
a sum of a square and a positive constant; positive derivative on an interval means strictly increasing, hence one to one
Find the input that produces 4
$$f(1)=1+2+1=4$$
small integers first: guessing is legitimate here because $f$ is increasing, so a hit is the only hit
$$a=f^{-1}(4)=1$$
that is the point where the rule wants the derivative
Independent numerical check: $f(1.02)=4.101208$, so the inverse takes $0.02$ of input for $0.101208$ of output, a slope of $0.1976$ near $y=4$ — that is $0.2$ to the accuracy this step size allows.
One guess, one derivative and one division; the cubic formula was never needed.
Read the two numbers in the right order and the rule is mechanical: the output says where to stand on the graph of $f$, and the slope found there gets flipped.
Why a zero slope leaves the inverse with no derivative at all
The theorem asks for $f'(a)\neq 0$, and the counterexample above is $x^{3}$ at the origin. A true or false item asks for more than one example: it asks whether the failure is guaranteed for every $f$. It is, and the argument is two lines.
Given
$f$ one to one and differentiable on an open interval $I$
$f(a)=b$ with $a$ in $I$
$f'(a)=0$
Find
whether $\left(f^{-1}\right)'(b)$ can exist
SolutionAssume it exists and let the chain rule finish the argument
$$f^{-1}\!\left(f(x)\right)=x\qquad\text{for every }x\in I$$
the defining property of the inverse, and an identity may be differentiated on both sides
the inverse relabels the same pair of points, and as $y\to b$ the matching $x\to a$, because a one to one continuous function on an interval is monotone and so is its inverse
the inner quotient tends to $f'(a)=0$ and is never $0$ itself, since $f$ is one to one; the reciprocal of something shrinking to zero grows past every bound
$$\text{graph of }f^{-1}:\ \text{vertical tangent at }b$$
a horizontal tangent reflects across the line $y=x$ into a vertical one, which is the picture of a difference quotient running off to infinity
Test the general claim on the case that has a formula
the quotient is not settling on a number; it multiplies by $100$ every time $y$ shrinks by a factor of $1000$
Answer $$\boxed{f'(a)=0\ \Longrightarrow\ \left(f^{-1}\right)'(b)\ \text{does not exist}}$$
Check
A check that uses neither argument and no formula for the inverse: take $g(x)=x^{3}+x^{5}$, again one to one with $g'(0)=0$. Here $g\!\left(10^{-3}\right)\approx 10^{-9}$, so the inverse moves $10^{-3}$ of output for $10^{-9}$ of input, a quotient near $10^{6}$; at $x=10^{-4}$ the same reading gives about $10^{8}$.
Three lines for the general claim, and the example is only a check afterwards.
The exam sentence is one line: differentiate $f^{-1}\!\left(f(x)\right)=x$ and read $\left(f^{-1}\right)'(b)\cdot f'(a)=1$. A counterexample answers the one claim it is an example of; this answers the claim about every $f$, which is what a true or false item is asking for.
Checkpoint
§12.1 — slope of an inverse from a single value●●○○○
Everything you know about $f$ is three lines of data. Nobody has written a formula for the inverse, and you will not need one.
Given
$f$ is differentiable and increasing on all of $\mathbb{R}$
$f(3)=7$
$f'(3)=2$
Find
(a) Compute $g'(7)$, where $g=f^{-1}$.
Hint 1/4
You are asked for a slope of the inverse at the output $7$. First find which input of $f$ produced it; that is where the work happens.
Hint 2/4
The rule is $\left(f^{-1}\right)'(b)=\dfrac{1}{f'\!\left(f^{-1}(b)\right)}$.
Hint 3/4
Here $b=7$, and $f(3)=7$ means $f^{-1}(7)=3$. The data once more: $f(3)=7$ and $f'(3)=2$.
Hint 4/4
The answer is $g'(7)=\tfrac{1}{2}$.
Show solutionFind the input behind the output
$$f(3)=7 \;\Longrightarrow\; f^{-1}(7)=3$$
the inverse reads the same pair of numbers backwards
Apply the rule at that input
$$g'(7)=\frac{1}{f'(3)}$$
the rule evaluates $f'$ at $f^{-1}(7)=3$, not at $7$
$$=\frac{1}{2}$$
the given value $f'(3)=2$
Answer $$\boxed{g'(7)=\tfrac{1}{2}}$$
Check
$f$ is increasing, so $g$ is too and the answer had to be positive; $f$ climbs $2$ units of output per unit of input near $x=3$, so $g$ climbs half a unit the other way.
Nothing here needed a formula for $g$. That is the whole point of the rule: one value and one derivative of $f$ are enough.
⚠ Evaluating the derivative at the output instead of the input
The number $b$ is the one in the question, so the eye puts it in every slot it fits.
12.2Bases other than e, and the slope that names them
Rewrites every base through $e$ and $\ln$: $a^{u}$ picks up $\ln a$, and $e^{u}$ keeps only $u'$ because $\ln e=1$.
Before spending the reciprocal-slope theorem on trigonometry, spend it on what the last section left unfinished: powers and logarithms with a base that is not $e$.
DefinitionDefinition 12.2: general exponential and logarithm
Conditions
$a>0$
$a\neq 1$ for the logarithm
$x>0$ for $\log_a x$ and for a real power $x^{n}$
Widen the statement: the exponent on an exam is almost never a bare $x$, so the working form is $\left(a^{u}\right)'=a^{u}u'\ln a$, and the same rule is the tool inside a tangency condition, where two curves are made to share a value and a slope, not only inside a plain differentiate this instruction. Widen the domain line as well: $x^{p}$ was restricted to $x>0$ only because $e^{p\ln x}$ needs it, while $x^{p/q}$ with $q$ odd is defined for negative $x$ too, and that branch has to be counted when a question asks how many solutions an equation has.
Name the natural base here instead of leaving it to the previous section: the case $a=e$ has $\ln e=1$, so $\dfrac{d}{dx}e^{u}=e^{u}u'$ — the exponential survives untouched and only the inner derivative is left behind. Read the same line from right to left and it is the substitution an exam hands you, $\int e^{u}u'\,dx=e^{u}+C$: the factor standing next to $e^{u}$ is exactly the $u'$ the chain rule would have produced.
Every base is the natural base in disguise: $a$ to a power is $e$ to that power times $\ln a$, and each derivative differs from the natural one by that constant — multiplying for the exponential, dividing for the logarithm.
Both derivatives in two lines
$\dfrac{d}{dx}e^{x\ln a}=e^{x\ln a}\ln a=a^{x}\ln a$ by the chain rule, the inner derivative being the constant $\ln a$. And $\log_a x$ is $\dfrac{1}{\ln a}$ times $\ln x$, so its derivative is $\dfrac{1}{\ln a}\cdot\dfrac{1}{x}$.
Every exponential passes through the same point, so what separates the bases is the steepness there — and that steepness is exactly $\ln a$. Base $e$ is the one with tangent slope $1$.
Looks like this, but is not
$\log_2$ is just $\ln$ with a different label on the base, so its derivative should also be $\dfrac{1}{x}$.
$\log_2 x=\dfrac{\ln x}{\ln 2}$ with $\ln 2\approx 0.693$, so the derivative is $\dfrac{1.443}{x}$ — about $44\%$ steeper than $1/x$ everywhere. A rescaled function has a rescaled slope.
shape
which rule
derivative
$x^{n}$, $n$ constant
power rule, valid for every real $n$ when $x>0$
$nx^{n-1}$
$a^{x}$, $a$ constant
general exponential
$a^{x}\ln a$
$u(x)^{v(x)}$
neither: take $\ln$ of both sides first
found case by case
Read the middle column before the third: the position of the variable, not the look of the expression, decides which line you are on.
Derivative of 3 to the power x squared
A constant base with a composite exponent: two rules meet in one line.
Given
$y=3^{x^{2}}$
Find
$y'$
SolutionName the outer and the inner function
$$y=3^{u},\quad u=x^{2}$$
the outer function is a constant to a power, the inner one is the exponent
Differentiate outer times inner
$$\frac{dy}{du}=3^{u}\ln 3$$
the general base rule
$$\frac{du}{dx}=2x$$
power rule on the exponent
$$y'=3^{x^{2}}\ln 3\cdot 2x$$
chain rule, then put $u=x^{2}$ back
Answer $$\boxed{y'=2x\,3^{x^{2}}\ln 3}$$
Check
Independent check by symmetry: $3^{x^{2}}$ is even, and an even function differentiable at $0$ has derivative $0$ there. The formula gives $2\cdot 0\cdot 1\cdot\ln 3=0$.
A constant base never leaves; it only picks up the factor $\ln a$.
Where the number e comes from as a limit
The definition $a^{x}=e^{x\ln a}$ takes $e$ for granted. Here is $e$ produced by a derivative — the exact statement the opening calculator experiment was groping at.
Given
$f(x)=\ln x$
$f'(1)=1$
Find
$\lim_{h\to 0}(1+h)^{1/h}$
SolutionWrite the known derivative as its difference quotient
$$1=f'(1)=\lim_{h\to 0}\frac{\ln(1+h)-\ln 1}{h}$$
the derivative of $\ln$ at $1$ is $1/1$, and it is also a limit by definition
$$=\lim_{h\to 0}\frac{1}{h}\ln(1+h)$$
$\ln 1=0$
Move the factor inside the logarithm
$$=\lim_{h\to 0}\ln\left[(1+h)^{1/h}\right]$$
a coefficient in front of a logarithm is an exponent inside it
Undo the logarithm
$$\lim_{h\to 0}(1+h)^{1/h}=e^{1}=e$$
$\exp$ is continuous, so it may be moved through the limit
Answer $$\boxed{\lim_{h\to 0}(1+h)^{1/h}=e\quad\text{and, with }h=\tfrac{1}{n},\quad \lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^{n}=e}$$
Check
Numerical check at $n=1000$: $(1.001)^{1000}=2.71692$ against $e=2.71828$. The gap shrinks like $1/n$, which is why the calculator needed a large exponent.
The base creeping to $1$ and the exponent running to infinity fight, and neither wins outright. That fight has a name and a method, both later in this section.
The natural base, forwards as a derivative and backwards as an integral
The box says a constant base leaves the factor $\ln a$ behind. Put $a=e$ and watch what that factor becomes; then read the same line from right to left, which is the form a substitution question hands you.
Two checks that do not repeat the rule. Symmetry: $e^{x^{2}/2}$ is even and differentiable, so its slope at $0$ has to be $0$, and the formula returns $0\cdot e^{0}=0$. Numbers: a symmetric difference quotient at $x=1$ with step $0.01$ gives $1.64883$, against $1\cdot e^{1/2}=1.64872$.
One chain rule, and the integral came free because it is the same line reversed.
A missing $u'$ and a missing $\ln a$ are the two places marks disappear in this family. With the natural base the second one is invisible, since $\ln e=1$, which is why it is safe to forget there and expensive to forget with any other base.
Which is larger, e to the pi or pi to the e
Two numbers are built from the same two ingredients with the roles swapped: $e^{\pi}$ and $\pi^{e}$. Decide which is larger, with no calculator. Then decide the same for $e^{\pi/2}$ and $\left(\tfrac{\pi}{2}\right)^{e}$, where the base has now dropped below $e$. One argument must settle both, and no function is handed to you.
Solution
The statement gives two constants and no function, so the student cannot start until the auxiliary function is built, which is exactly the reflex the note never triggers.
Turn a comparison of two numbers into a comparison a derivative can reach
$\ln$ is strictly increasing, so the inequality survives it in both directions and nothing is assumed by taking it; it is also the only move that pulls an exponent down into a factor, and a factor is the only thing calculus can differentiate
dividing by $e\pi>0$ keeps the direction, but the point is not the direction: it leaves each side depending on one of the two numbers alone, which is what makes a single function of a single variable appear
Recognise both sides as values of one function
$$\frac{1}{e}=\frac{\ln e}{e}$$
writing the loose constant as a value of the same expression is what converts the question from a comparison of two unrelated numbers into a comparison of one function at two points
now the whole problem is the shape of one graph, so the standard maximum machinery of this course applies to a question that never mentioned a function
Locate the peak of g and make the conclusion global
$\ln t<1$ exactly for $t<e$, so the sign changes once and only once; a single change from plus to minus is what upgrades a critical point to an absolute maximum over the whole domain, and the absolute version is needed because $\pi$ is nowhere near $e$
$$g(t)<g(e)=\frac{1}{e}\qquad\text{for every }t>0,\ t\neq e$$
$g$ is strictly monotone on each side of $e$ since $g'$ vanishes only there, so the inequality is strict, which matters: a non-strict version would leave the comparison undecided
the theorem never asked which side of $e$ the number sits on, only that it is not $e$; the maximum is global, so a base below $e$ loses in the same direction as one above it
the larger base won in one case and lost in the other, so "bigger base wins" is not a rule; that picture compares $a^{x}$ and $b^{x}$ at a shared exponent, and here the two numbers share no exponent at all
this, and only this, is what decides such a swap; the sizes of $a$ and $b$ enter solely through where they sit relative to the peak at $e$
Answer $$\boxed{e^{t}>t^{e}\ \text{for every }t>0\text{ with }t\neq e;\quad\text{in particular }e^{\pi}>\pi^{e}\ \text{and}\ e^{\pi/2}>\left(\tfrac{\pi}{2}\right)^{e}}$$
Check
Numbers, as a check after the fact and not as the proof: $e^{\pi}=23.141$ against $\pi^{e}=22.459$, and $e^{\pi/2}=4.810$ against $(\pi/2)^{e}=3.413$. Structural check on the exclusion $t\neq e$: at $t=e$ the claim degenerates to $e^{e}=e^{e}$, which is precisely what a strict maximum attained at $t=e$ must produce, so the excluded point is the equality case and not a technicality.
A limit whose entire answer is the sign of one constant
Evaluate $$L=\lim_{x\to\infty}x^{\left(3+\frac{1}{x}\right)^{\pi}-\pi^{\,3+\frac{1}{x}}}.$$ Note before starting: nothing here is a $0/0$ or an $\infty/\infty$ quotient, and neither $3$ nor $\pi$ equals $e$.
Solution
This is the shape the papers actually use, where the constant comparison is buried as the deciding sub-step of a limit and both numbers sit on the same side of the peak, so the maximum value is useless and only monotonicity settles it.
naming the exponent as one object separates what still moves from the base, and blocks the reflex of taking logarithms of the whole expression before knowing whether anything is indeterminate at all
$$\lim_{x\to\infty}u(x)=3^{\pi}-\pi^{3}=:c$$
$s\mapsto s^{\pi}=e^{\pi\ln s}$ and $s\mapsto\pi^{s}=e^{s\ln\pi}$ are continuous at $s=3$, so this limit is plain substitution and costs nothing; the rewrite through $e$ is what makes both continuous in the first place
of the power shapes only $\infty^{0}$ is indeterminate; with $c\neq 0$ the answer is already forced to be $0$ or $\infty$, so l'Hopital has nothing to act on and the whole exercise reduces to the sign of $c$
Settle the sign by comparing two numbers whose roles swap
$$c>0\iff 3^{\pi}>\pi^{3}\iff \pi\ln 3>3\ln\pi$$
$\ln$ is strictly increasing so the comparison passes through it untouched, and only after it does each side become a product rather than a tower
neither $3$ nor $\pi$ is $e$, so the peak value $1/e$ answers nothing here; what does the work is that both numbers lie on the falling side, and the sign of $g'$ is carried entirely by $1-\ln t$
strict decrease turns the order of the inputs into the reversed order of the outputs, and the chain above was equivalences throughout, so the conclusion travels back to the powers
Feed the sign back into the limit
$$x^{u(x)}=e^{u(x)\ln x}\qquad (x>0)$$
a moving exponent is readable only through $e$ and $\ln$; this is the same rewrite that defined a general power, used here on an exponent that is not constant
$u$ is not the constant $c$, and replacing it by $c$ would be assuming what has to be shown; the definition of $\lim u=c$ supplies a fixed positive floor, which is all the argument needs
a positive floor times an unbounded factor is unbounded, and $\exp$ is increasing and unbounded, so the exponent running to $+\infty$ drags the whole expression there
The two constants are close: $\ln 3/3=0.36620$ against $\ln\pi/\pi=0.36438$, a gap under $0.002$, which is exactly why decimals may check the answer but cannot be the proof. Consistency with the verdict: at $x=10$ the exponent is $0.200$ and $x^{u(x)}=1.586$; at $x=10^{3}$ the exponent is $0.536$ and $x^{u(x)}=40.4$; at $x=10^{6}$ it is $1690$. Slow, but climbing, so the answer is the $\infty$ branch and not the $0$ branch.
Checkpoint
§12.2 — differentiating a constant to a variable power●●○○○
Thirty seconds. The base is a constant and the exponent is the variable, which is the opposite arrangement from the power rule.
Given
$y=5^{x}$
Find
(a) Which expression is $\dfrac{dy}{dx}$?
Hint 1/4
Decide first which of the two positions the variable occupies: base or exponent. That decision picks the rule; nothing else does.
Hint 2/4
For a constant base, $\dfrac{d}{dx}a^{x}=a^{x}\ln a$, because $a^{x}=e^{x\ln a}$.
Hint 3/4
Here $a=5$, so the constant that the chain rule pulls out is $\ln 5$, and the function $5^{x}$ itself survives untouched.
Hint 4/4
The derivative is $5^{x}\ln 5$.
Show solutionRewrite through the exponential we can differentiate
$$5^{x}=e^{x\ln 5}$$
$\ln 5$ is a constant, so this is $e$ to a constant times $x$
$$\frac{d}{dx}e^{x\ln 5}=e^{x\ln 5}\cdot\ln 5$$
chain rule; the inner derivative is the constant $\ln 5$
$$=5^{x}\ln 5$$
translate back
Answer $$\boxed{5^{x}\ln 5}$$
Check
Sanity check on size: $\ln 5\approx 1.609>1$, so $5^{x}$ grows faster than $e^{x}$ does at the same height, which is what a base larger than $e$ should do.
⚠ Using the power rule on a constant base
The shape $\text{something}^{\text{something}}$ triggers the most practised rule.
wrong$$\frac{d}{dx}2^{x}=x\,2^{x-1}$$
right$$\frac{d}{dx}2^{x}=2^{x}\ln 2$$
⚠ Putting the constant on the wrong side of the fraction
Both formulas contain $\ln a$, and only the position distinguishes them.
wrong$$\frac{d}{dx}\log_a x=\frac{\ln a}{x}$$
right$$\frac{d}{dx}\log_a x=\frac{1}{x\ln a}$$
12.3Restricting the sine until it can be inverted
The two derivatives, plus $\arctan x\to\pm\pi/2$, the bound $\left|\arctan x\right|<\pi/2$, and $\int\frac{dx}{1+x^{2}}=\arctan x+C$.
The theorem needs a one-to-one function, and no trigonometric function is one to one. So the first move is not calculus but surgery on the domain.
TheoremTheorem 12.3: the two derivatives you will use most
Conditions
$\arcsin:[-1,1]\to\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, the branch on which the sine increases
the arcsine formula holds on the open interval $|x|<1$ only
Add to the conditions: $\arctan x\to\pi/2$ as $x\to+\infty$ and $\arctan x\to-\pi/2$ as $x\to-\infty$, so $\left|\arctan x\right|\lt \pi/2$ for every real $x$ and an arctangent inside a limit is a bounded quantity that freezes to a constant. Read the same formula backwards and it is an antiderivative, $\int dx/(1+x^{2})=\arctan x+C$, so $\int_{0}^{x}dt/(1+t^{2})\to\pi/2$ is a numerator that stays finite rather than one that blows up.
The same two values hold through a composition: if $u(x)\to+\infty$ then $\arctan u(x)\to\pi/2$, and if $u(x)\to-\infty$ then $\arctan u(x)\to-\pi/2$, because a limit passes through a continuous outer function — the rule stated with continuity earlier in the course. So $\arctan\!\left(\sqrt3\,t\right)$ and $\arctan\!\left(e^{t}\right)$ both freeze at $\pi/2$ as $t\to\infty$, and neither of them has to be differentiated to see it.
The arcsine climbs at one over the cosine of the angle it returns, and on its branch that cosine is $\sqrt{1-x^{2}}$; the arctangent climbs at $1/(1+x^{2})$, never zero, so it is defined for every real number.
Both, by implicit differentiation
Put $y=\arcsin x$, so $\sin y=x$ with $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Differentiating both sides with respect to $x$ gives $\cos y\cdot y'=1$, hence $y'=1/\cos y$. On that branch $\cos y\ge 0$, so $\cos y=+\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}$ — the sign is decided by the branch, not by taste. For the arctangent, $\tan y=x$ gives $\sec^{2}y\cdot y'=1$ and $\sec^{2}y=1+\tan^{2}y=1+x^{2}$.
Why an algebraic root appears in the derivative of a trigonometric inverse: write $\theta=\arcsin \textcolor{#1f6feb}{x}$ into a right triangle with hypotenuse $1$, and $\cos\theta$ is read off as $\textcolor{#d1690a}{\sqrt{1-x^{2}}}$.
Looks like this, but is not
The notation $\sin^{-1}x$ has an exponent $-1$ in it, and $u^{-1}$ means $1/u$ everywhere else in algebra, so $\sin^{-1}\left(\tfrac12\right)$ should be $2$.
$\sin^{-1}\left(\tfrac12\right)=\tfrac{\pi}{6}\approx 0.524$, an angle, while $\dfrac{1}{\sin\left(\tfrac12\right)}\approx 2.086$, a ratio. In this notation $-1$ means inverse function; the reciprocal is $\csc$.
Derivative of arctan of e to the x
A composite whose inner function is itself an exponential; two known rules meet.
Given
$y=\arctan\left(e^{x}\right)$
Find
$y'$
SolutionSplit into outer and inner
$$y=\arctan u,\quad u=e^{x}$$
the arctangent is applied to something, so it is the outer function
Two independent checks. Sign: $\arctan$ and $e^{x}$ both increase, so the derivative must be positive, and it is a quotient of positive numbers. End behaviour: as $x\to\infty$ the formula tends to $0$, matching the flattening curve.
Inverse trigonometric functions are ordinary outer functions; nothing about them changes the chain rule.
The combination whose derivative collapses to arcsine
This one looks heavier than it is, and the payoff is a result you will meet again when integrating.
the two awkward terms are exact negatives, which is the reason this combination is worth remembering
Answer $$\boxed{F'(x)=\arcsin x}$$
Check
Numerical check at $x=\tfrac12$, where the answer should be $\arcsin 0.5=0.523599$: a symmetric difference quotient with step $0.01$ gives $\dfrac{F(0.51)-F(0.49)}{0.02}=0.523612$, agreeing to four decimals.
Product rule once, chain rule once, and the whole middle of the answer cancels.
Read backwards this says that $x\arcsin x+\sqrt{1-x^{2}}$ is an antiderivative of $\arcsin$, which is how the next chapter will use it.
Checkpoint
§12.3 — the arcsine derivative at a point●●○○○
Thirty seconds, no chain rule involved. The formula is quoted at a single point well inside the domain.
Given
$y=\arcsin x$
the point $x=\tfrac{1}{2}$
Find
(a) Evaluate $y'\left(\tfrac{1}{2}\right)$ and leave the answer exact.
Hint 1/4
You are asked for a number, not a function: quote the derivative formula and then substitute the one value of $x$.
Hint 2/4
On the principal branch, $(\arcsin x)'=\dfrac{1}{\sqrt{1-x^{2}}}$ for $|x|<1$.
Hint 3/4
With $x=\tfrac{1}{2}$ the square inside is $\tfrac14$, so the root is $\sqrt{1-\tfrac14}=\sqrt{\tfrac34}$.
Hint 4/4
The value is $\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}$.
Each pair of co-functions adds up to a right angle, so their derivatives add up to zero: whatever one of them does, its partner does with the opposite sign.
Why the sum is a right angle, and what that costs
$\cos\left(\tfrac{\pi}{2}-\theta\right)=\sin\theta$, and if $\theta=\arcsin x$ lies in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ then $\tfrac{\pi}{2}-\theta$ lies in $[0,\pi]$ — the branch $\arccos$ returns. So $\arccos x=\tfrac{\pi}{2}-\arcsin x$; differentiate and the minus sign appears by itself. The secant case differs: books disagree on the branch for negative $x$, so that sign is a convention. For $x>1$, $\sec y=x$ gives $\sec y\tan y\cdot y'=1$ with $\tan y=\sqrt{x^{2}-1}$, hence $y'=1/\left(x\sqrt{x^{2}-1}\right)$.
The identity as a picture: reflecting the arcsine graph in the horizontal line $y=\pi/4$ produces the arccosine graph, and a reflection reverses every slope.
Looks like this, but is not
The two graphs differ by the constant $\tfrac{\pi}{2}$, and adding a constant changes no slope. So $\arcsin$ and $\arccos$ should have equal derivatives.
The formula is $\arccos x=\tfrac{\pi}{2}-\arcsin x$, not $\tfrac{\pi}{2}+\arcsin x$: the minus flips the graph before the constant slides it. At $x=\tfrac12$ the slopes are $+1.155$ and $-1.155$.
function
domain
principal values
derivative
$\arcsin x$
$[-1,1]$
$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\dfrac{1}{\sqrt{1-x^{2}}}$
$\arctan x$
$\mathbb{R}$
$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\dfrac{1}{1+x^{2}}$
$\operatorname{arcsec} x$
$\vert x\vert\ge 1$
stated by the question
$\dfrac{1}{x\sqrt{x^{2}-1}}$ for $x>1$
$\arccos x$
$[-1,1]$
$[0,\pi]$
$-\dfrac{1}{\sqrt{1-x^{2}}}$
$\operatorname{arccot} x$
$\mathbb{R}$
$(0,\pi)$
$-\dfrac{1}{1+x^{2}}$
$\operatorname{arccsc} x$
$\vert x\vert\ge 1$
stated by the question
$-\dfrac{1}{x\sqrt{x^{2}-1}}$ for $x>1$
The derivative column holds on the open part of the domain: the arcsine, arccosine and both secant rows lose it at $x=\pm 1$, where the tangent turns vertical, while $1+x^{2}$ never vanishes.
Proving the identity instead of quoting it
Suppose you cannot remember whether the sum is $\tfrac{\pi}{2}$ or $\pi$. Calculus settles it in two lines, for any suspected identity.
Independent check at a different point: $x=1$ gives $\tfrac{\pi}{2}+0=\tfrac{\pi}{2}$, and $x=-1$ gives $-\tfrac{\pi}{2}+\pi=\tfrac{\pi}{2}$. Two more agreements, neither used in the proof.
Zero derivative plus one evaluated point is a complete proof of an identity. Remember the technique rather than the six formulas.
Checkpoint
§12.4 — what a constant sum forces on two derivatives●●○○○
A single sentence to judge. It is the kind of claim that sounds harmless because the identity quoted inside it is true.
Given
$\arcsin x+\arccos x=\dfrac{\pi}{2}$ for every $x$ in $[-1,1]$
Find
(a) True or false: because the two functions differ only by the constant $\tfrac{\pi}{2}$, they have the same derivative.
Hint 1/4
Rewrite the claim as an equation between the two functions and look at what kind of operation turns one into the other.
Hint 2/4
If $u+v$ is constant then $u'+v'=0$, so $v'=-u'$: the derivatives are opposite, not equal.
Hint 3/4
Here $\arccos x=\tfrac{\pi}{2}-\arcsin x$; the arcsine is not shifted, it is shifted and flipped.
Hint 4/4
The statement is false: the derivatives are negatives of one another.
an identity may be differentiated on both sides; a constant differentiates to zero
$$(\arcsin x)'+(\arccos x)'=0$$
the derivative of a sum
$$(\arccos x)'=-\frac{1}{\sqrt{1-x^{2}}}$$
move one term across
Answer $$\boxed{(\arccos x)'=-(\arcsin x)'}$$
Check
Check against the graph: $\arccos$ falls from $\pi$ to $0$ as $x$ runs from $-1$ to $1$, so its derivative must be negative everywhere; the formula is.
⚠ Copying the arcsine root into the arcsecant formula
Both formulas contain a square root and a difference of squares, and the order of the two terms is easy to swap.
right$$x>1:\ \frac{1}{x\sqrt{x^{2}-1}};\qquad x<-1:\ \text{sign fixed by the branch the question states}$$
12.5The exponential pair that behaves like sine and cosine
Builds $\cosh$ and $\sinh$ out of $e^{x}$; reach for them when an identity or inverse must read as a logarithm.
The section title names one more family. It is built from $e^{x}$ with no new machinery, and its inverses are the one place a closed form can be checked by hand.
DefinitionDefinition 12.5: the hyperbolic functions
Split $e^{x}$ into its even and odd halves: the even half is $\cosh$, the odd half $\sinh$, each is the derivative of the other with no minus sign, and the difference of their squares is one.
The identity in one line
$\left(e^{x}+e^{-x}\right)^{2}-\left(e^{x}-e^{-x}\right)^{2}=4e^{x}e^{-x}=4$; divide both sides by $4$. The same computation with a plus sign gives $\cosh(2x)$, which is certainly not constant.
Where the name comes from: $(\cos t,\sin t)$ runs around a circle, and $(\cosh t,\sinh t)$ runs along one branch of a hyperbola, because the identity it satisfies has a minus sign.
Looks like this, but is not
Near the origin $\cosh x$ matches the parabola $1+\tfrac{x^{2}}{2}$ — same value, slope and curvature at $x=0$ — so it is presumably a parabola in disguise.
At $x=5$ the parabola gives $13.5$ while $\cosh 5=74.21$. A parabola grows like a power, $\cosh$ like $e^{x}/2$; agreement near one point says nothing about the far field.
function
definition or closed form
derivative
$\sinh x$
$\dfrac{e^{x}-e^{-x}}{2}$
$\cosh x$
$\cosh x$
$\dfrac{e^{x}+e^{-x}}{2}$
$\sinh x$
$\tanh x$
$\dfrac{\sinh x}{\cosh x}$
$\operatorname{sech}^{2}x=\dfrac{1}{\cosh^{2}x}$
$\sinh^{-1}x$
$\ln\left(x+\sqrt{x^{2}+1}\right)$, all $x$
$\dfrac{1}{\sqrt{1+x^{2}}}$
$\cosh^{-1}x$
$\ln\left(x+\sqrt{x^{2}-1}\right)$, $x\ge 1$
$\dfrac{1}{\sqrt{x^{2}-1}}$, $x>1$
$\tanh^{-1}x$
$\tfrac12\ln\dfrac{1+x}{1-x}$, $\vert x\vert<1$
$\dfrac{1}{1-x^{2}}$
The last three rows are logarithms, so column three could be obtained by differentiating column two; the reciprocal-slope theorem is faster.
Solving x equals sinh y to get the logarithmic form of the inverse
Every inverse so far had to stay unwritten. This one can be solved for, because the equation is secretly a quadratic.
Given
$x=\sinh y=\dfrac{e^{y}-e^{-y}}{2}$
Find
$y=\sinh^{-1}x$ in closed form
SolutionTurn it into a quadratic
$$2x=e^{y}-e^{-y}$$
clear the denominator
$$2xe^{y}=e^{2y}-1$$
multiply through by $e^{y}$, which is never zero so nothing is lost
$$t^{2}-2xt-1=0,\quad t=e^{y}$$
naming $t=e^{y}$ makes the quadratic visible
Solve and discard the impossible root
$$t=x\pm\sqrt{x^{2}+1}$$
quadratic formula with $a=1$, $b=-2x$, $c=-1$
$$t=x+\sqrt{x^{2}+1}$$
$t=e^{y}>0$, and $\sqrt{x^{2}+1}>\vert x\vert$ makes the minus root negative
Exact check at $x=\tfrac34$: the formula gives $\ln\left(0.75+\sqrt{1.5625}\right)=\ln(0.75+1.25)=\ln 2$, and running it forward, $\sinh(\ln 2)=\tfrac{2-\tfrac12}{2}=\tfrac34$. The two directions agree.
Differentiating this closed form gives $\left(\sinh^{-1}x\right)'=\dfrac{1}{\sqrt{1+x^{2}}}$, which is what the reciprocal-slope theorem gives through $1/\cosh y$. With an inner function the chain rule applies as usual: $\left(\sinh^{-1}(3x)\right)'=\dfrac{3}{\sqrt{1+9x^{2}}}$.
Checkpoint
§12.5 — the hyperbolic identity, sign and all●●○○○
The trigonometric identity next door is $\sin^{2}+\cos^{2}=1$, and the habit of writing a plus sign there is strong.
Given
$\cosh x=\dfrac{e^{x}+e^{-x}}{2}$
$\sinh x=\dfrac{e^{x}-e^{-x}}{2}$
Find
(a) True or false: $\cosh^{2}x+\sinh^{2}x=1$ for every real $x$.
Hint 1/4
Do not test the claim at $x=0$ only; that point makes several different claims look true. Pick a second value as well.
Hint 2/4
Expanding the two definitions gives $\cosh^{2}x-\sinh^{2}x=1$: the identity has a minus sign, and $\cosh^{2}x+\sinh^{2}x$ is $\cosh(2x)$ instead.
Hint 3/4
At $x=1$: $\cosh 1\approx 1.5431$ and $\sinh 1\approx 1.1752$, so the sum of squares is about $2.381+1.381=3.762$.
Hint 4/4
False: the sum of squares is $\cosh(2x)$, and only the difference is $1$.
one counterexample is enough to kill a claim made for every $x$
$$\cosh^{2}1-\sinh^{2}1\approx 2.381-1.381=1$$
the same two numbers satisfy the identity with the minus sign
Answer $$\boxed{\cosh^{2}x-\sinh^{2}x=1}$$
Check
Algebraic confirmation, independent of the decimals: $\left(e^{x}+e^{-x}\right)^{2}-\left(e^{x}-e^{-x}\right)^{2}=4$, and dividing by $4$ gives exactly $1$.
⚠ Importing the minus sign from the circular derivative
$(\cos x)'=-\sin x$ is drilled far harder than its hyperbolic neighbour.
wrong$$(\cosh x)'=-\sinh x$$
right$$(\cosh x)'=+\sinh x$$
⚠ Writing the identity with a plus sign
The Pythagorean identity has a plus, and the two families look alike on paper.
wrong$$\cosh^{2}x+\sinh^{2}x=1$$
right$$\cosh^{2}x-\sinh^{2}x=1$$
12.6l'Hôpital's rule and the two shapes it accepts
Replaces a collapsing quotient by the ratio of derivatives once substitution shows $0/0$ or $\infty/\infty$, repeated once per order of vanishing.
Every family here produces limits substitution refuses to answer: the opening calculator experiment, and any quotient in which both parts vanish. One theorem covers them, with hypotheses worth obeying.
TheoremTheorem 12.6: l'Hôpital's rule
Conditions
$f$ and $g$ are differentiable near $a$, except possibly at $a$ itself
$g'(x)\neq 0$ near $a$
either $f(x)\to 0$ and $g(x)\to 0$, or $|f(x)|\to\infty$ and $|g(x)|\to\infty$
the limit on the right exists or is $\pm\infty$
$a$ may be a number, a one-sided approach, or $\pm\infty$
Widen the last condition: if $\lim f'/g'$ fails to exist, the rule is silent and silence is not an answer, so the original limit may still exist and must be found from the original expression; and if the quotient of derivatives reproduces the shape it came from, as $\sqrt{x^{2}+1}/x$ and $(1+x^{3})^{1/3}/x$ do, the rule is legal but never terminates. Add too that the shape is read at the target point as printed, so a bounded numerator, including a convergent definite integral, over something blowing up is not $\infty/\infty$ at all.
When both parts of a quotient collapse together, compare how fast each one is collapsing instead: the ratio of the speeds answers the question that the ratio of the sizes could not.
Why it is true in the simplest case
Suppose $f(a)=g(a)=0$, both differentiable at $a$ with $g'(a)\neq 0$. Divide top and bottom by $x-a$: $$\dfrac{f(x)}{g(x)}=\dfrac{\left(f(x)-f(a)\right)/(x-a)}{\left(g(x)-g(a)\right)/(x-a)}\longrightarrow\dfrac{f'(a)}{g'(a)}$$ — nothing but the definition of the derivative. The general statement, with $a$ infinite or the parts blowing up, needs the two-function Mean Value Theorem and is proved in the book.
The reason the rule works, drawn: near a common zero each curve is almost its own tangent line, and two lines through one point have heights in the ratio of their slopes.
Looks like this, but is not
$$\displaystyle\lim_{x\to 0}\frac{x+1}{x^{2}+2}$$ is a quotient of differentiable functions, so differentiating top and bottom should give $\dfrac{1}{2x}$, which runs off to infinity.
Substitution gives $\tfrac12$, so there was nothing to resolve. The rule is not a technique for quotients but for two specific collapses. Used outside them it returns a confident wrong answer instead of an error.
The limit that needs the rule twice
A quotient where one application is not enough — and where the form has to be checked again before the second.
the second hypothesis asks for a nonzero bottom derivative near the target, not at it, and a punctured neighbourhood is all the rule needs; keys pay for this line separately from the answer
Differentiate top and bottom, separately
$$L=\lim_{x\to 0}\frac{e^{x}-1}{2x}$$
each part differentiated on its own; this is not the quotient rule
$$\frac{e^{0}-1}{0}=\frac{0}{0}$$
check again: still indeterminate, so a second application is legal
Apply it once more and read off the answer
$$L=\lim_{x\to 0}\frac{e^{x}}{2}=\frac{1}{2}$$
now substitution works, which is the signal to stop
Answer $$\boxed{L=\tfrac12}$$
Check
Numerical check at $x=0.1$: $\dfrac{e^{0.1}-1-0.1}{0.01}=\dfrac{0.00517092}{0.01}=0.517$, and the values keep falling towards $0.5$ as $x$ shrinks.
Two applications, and one form check between them; the check is the step people skip.
Every application needs its own permission. Had the second check come out $0/2$, the answer was already $0$ and differentiating again would have been a mistake.
How an exponential beats a polynomial
The other accepted shape, infinity over infinity, in the comparison that decides most end-behaviour questions.
both parts grow without bound, which is the second accepted shape
$$g(x)=e^{x},\qquad g'(x)=e^{x}\neq 0\ \text{for every }x$$
the second hypothesis is not only for the $\tfrac{0}{0}$ branch: the $\tfrac{\infty}{\infty}$ branch asks for the same nonzero bottom derivative near the target, and here it is nonzero everywhere, so the line costs one second to write and the application is not legitimate without it
Differentiate until the top runs out
$$L=\lim_{x\to\infty}\frac{2x}{e^{x}}$$
first application; still $\infty/\infty$
$$L=\lim_{x\to\infty}\frac{2}{e^{x}}=0$$
second application; now the top is a constant and the bottom still grows, so the shape is $2/\infty$ and the limit is $0$
Answer $$\boxed{L=0}$$
Check
Numerical check at $x=20$: $\dfrac{400}{e^{20}}\approx 8.2\times 10^{-7}$, already a millionth, and each further unit of $x$ divides it by roughly $e$.
Two differentiations flattened the polynomial and left the exponential untouched. With $x^{n}$ it takes $n$ steps and ends the same way — the precise sense in which exponentials outrun polynomials.
A numerator that vanishes to fourth order, taken to the end
The example above stopped after two applications, and the method boxes warn you about reaching for a third. Here four are not a warning but the plan: the top vanishes to the same order as the bottom, so the number of applications is known before the first derivative is taken.
the other hypothesis: the bottom derivative vanishes only at the target itself, which the rule allows, and writing the line is a marked step
$$\text{top vanishes to order }4,\qquad \text{bottom}=x^{4}$$
each application takes one power off the bottom and one term off the top, so exactly four will finish it; announcing the number now is what keeps the third derivative from feeling like a wrong turn
Apply it four times, re-reading the shape between applications
the standard limit $(e^{x}-1)/x\to 1$ is on the formula card and may be quoted, so the fourth application is optional; the count of four refers to the rule alone
Answer $$\boxed{L=\dfrac{1}{24}}$$
Check
Numbers, from outside the method: the quotient is $0.042514$ at $x=0.1$ and $0.042087$ at $x=0.05$, closing on $1/24=0.041667$ from above. Consistency reading: four differentiations turn $x^{4}$ into $4!=24$ while the top tends to $e^{0}=1$, which is the fraction that came out.
Four applications, four shape checks, and the count was known from the first line.
Ask how deep the zero is, not how tired you are. A third or fourth derivative is a wrong turn only when the quotient stops getting simpler; when every application removes one power, the number of applications is a prediction you can write down before starting.
Three applications, or one identity and a standard limit
A limit both routes settle. Running the rule three times is legal and earns the marks; opening the shifted angle first reaches the same number in one line, and under exam time that is the difference worth having.
the addition formula with both $\cos(\pi/4)$ and $\sin(\pi/4)$ equal to $\tfrac{\sqrt2}{2}$; it is the one move that turns a sum of two functions into a single sine, and it comes back with the trigonometric integrals later in the course
the substitution is legitimate because $t\to 0$ exactly when $x\to-\pi/4$, and $\left(\sqrt2\right)^{3}=2\sqrt2$ is a constant that walks out of the limit
$$=2\sqrt2\cdot 1^{3}=2\sqrt2$$
the standard limit $\sin t/t\to 1$ is on the formula card and may be quoted; cubing is continuous, so the cube of the limit is the limit of the cube
third application; the bottom is now the constant $6$, so this is the last one and the shape check is over
$$=\frac{6\left(\sqrt2\right)^{3}-0}{6}=2\sqrt2$$
at the target $\cos x-\sin x=\sqrt2$ while $\sin x+\cos x=0$, so the second term drops and only the cube is left
Answer $$\boxed{L=2\sqrt2\approx 2.828}$$
Check
The two routes use different machinery and agree. Numerically, one hundredth to the right of the target the quotient is $2.82829$, against $2\sqrt2=2.82843$.
Route one: one identity, one substitution, one quoted limit. Route two: three applications, three shape checks, and two derivatives that run over a full line.
Both routes are worth the same marks, so the identity is not about being clever: it is about not spending the minutes on differentiating a cube three times when the rest of the paper needs them. Whenever $\sin x$ and $\cos x$ appear added, $\sqrt2\,\sin\!\left(x+\tfrac{\pi}{4}\right)$ is available.
A two sided zero of order three, differentiated on both sides of the origin
The target is $x\to 0$ from both sides and the denominator is $x^{3}$, so the left of the origin has to be differentiated as well as the right. The power rule card in this section carries the restriction $x>0$; the first job is to say why that restriction does not reach this limit, and the second is the usual pair of hypotheses.
the bottom derivative vanishes only at the target itself, which is what a punctured neighbourhood means and what the rule asks for; it is the second half of that licence, so this line is part of the argument and not decoration
$$\frac{d}{dx}x^{3}=3x^{2}\quad\text{for every real }x,\qquad\text{also }x<0$$
the condition $x>0$ on the real exponent card belongs to the derivation through $e^{n\ln x}$, which needs a positive base. For a whole number exponent the product and sum rules from the differentiation section give the same formula with no sign condition, so both one sided limits are covered by one calculation instead of by luck
Apply it once, then simplify before deciding anything else
one factor of $x$ cancels, legal because a limit never evaluates at $x=0$; simplifying before the next shape check is what keeps the count of applications honest
third application; the bottom is now the constant $6$, substitution returns a number and that is the signal to stop
Answer $$\boxed{L=\dfrac13}$$
Check
Two checks from outside the method. Parity: the numerator is odd and $x^{3}$ is odd, so the quotient is even and the two one sided limits are forced to agree — the quotient is $0.33300$ at $x=0.1$ and the same $0.33300$ at $x=-0.1$, which is the two sidedness checked with numbers rather than assumed. Independence: the short route quoted a standard limit and the long route never did, and both returned $1/3$.
One application plus one quoted limit, or three applications; the hypothesis lines cost two more.
A denominator of $x^{3}$ at a two sided target is not a reason to hesitate. A whole number power needs no sign condition, so the left of the origin is differentiated by the same formula as the right; the restriction on the card is a property of one derivation, not of the rule.
Checkpoint
§12.6 — deciding whether the rule is allowed●●●○○
Before differentiating anything, substitute. Only two shapes give the rule permission to start, and one of these four limits has one of them.
Given
l'Hôpital's rule applies to the shapes $\dfrac{0}{0}$ and $\dfrac{\infty}{\infty}$ only
Find
(a) To which limit may the rule be applied directly, with no rewriting first?
Hint 1/4
Substitute the limit point into each numerator and each denominator separately and write down the pair of numbers.
Hint 2/4
The rule needs both parts to go to zero, or both to blow up. A number over a number, a number over zero, zero over a number: none is its business.
Hint 3/4
The four substitutions give, in order: $0/0$; $1/0$; $0/1$; $0/4$.
Hint 4/4
Only $\dfrac{\cos x-1}{x^{2}}$ has an indeterminate shape at $x\to 0$.
Show solutionSubstitute in each
$$\frac{\cos 0-1}{0}=\frac{0}{0}$$
indeterminate: the rule may start here
$$\frac{\cos 0}{0}=\frac{1}{0}$$
a non-zero number over zero is not indeterminate; the size runs to infinity
$$\frac{0+0}{0+1}=\frac{0}{1}=0$$
already an answer
$$\frac{4-4}{2+2}=\frac{0}{4}=0$$
already an answer; the zero on top is not matched below
right$$\lim_{x\to 0}\frac{e^{x}}{2}=\frac12\quad\text{stop: the shape is no longer indeterminate}$$
12.7The five forms that need a rewrite before the rule is allowed
Converts products, differences and powers into a quotient the rule accepts; power forms need logarithms, so the answer ends exponential.
The rule accepts exactly two shapes. Everything else — the calculator experiment included — has to be pushed into one of them first, and there are only three pushes.
MethodMethod 12.7: the three rewrites
Conditions
$0\cdot\infty$: send one factor to the denominator
$\infty-\infty$: combine into a single fraction, or factor the dominant term out
$0^{0}$, $1^{\infty}$, $\infty^{0}$: take the logarithm, then handle the $0\cdot\infty$ that appears, and exponentiate at the end
A product becomes a quotient by moving one factor downstairs; a difference becomes a quotient over a common denominator; a power becomes a product by taking logarithms — which is why every power problem ends in an exponential.
Why the logarithm step is legal
$u^{v}=e^{v\ln u}$ for $u>0$ and $\exp$ is continuous, so the limit passes through it: if $v\ln u\to c$ then $u^{v}\to e^{c}$, and if $v\ln u\to-\infty$ then $u^{v}\to 0$. That continuity is also what makes the final exponentiation compulsory.
The whole decision in one chart: the left column is what substitution reported, the right column the single move that shape needs. Every move but the last sends you back to the top to substitute again.
Looks like this, but is not
$1$ raised to any power is $1$, so a limit of shape $1^{\infty}$ must be $1$ — settling the opening experiment without any work.
$\left(1+\tfrac{1}{n}\right)^{n}\to e=2.71828$ and $\left(1+\tfrac{3}{n}\right)^{n}\to e^{3}=20.0855$: same shape, different answers, neither $1$. The base is only heading to $1$, and how fast decides the outcome.
The calculator experiment from the first paragraph, finished exactly
The opening scene had base $1.001$ and exponent $1000$. Here is the general version, with a constant in the numerator so the mechanism is visible.
chain rule; the inner derivative of $3/x$ is $-3/x^{2}$
$$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^{2}}$$
the denominator
$$\ln L=\lim_{x\to\infty}\frac{3}{1+3/x}=3$$
the two factors $-1/x^{2}$ cancel, which is the reason $1/x$ was the right thing to divide by
Undo the logarithm
$$L=e^{3}\approx 20.0855$$
the step that turns $\ln L$ into $L$; leaving it out is the classic lost mark
Answer $$\boxed{L=e^{3}}$$
Check
Numerical check at $x=1000$: $(1.003)^{1000}=19.995$, within half a percent of $e^{3}=20.0855$. With $3$ replaced by $1$ it gives $(1.001)^{1000}=2.7169$ against $e=2.71828$ — the number in the first paragraph.
One logarithm, one rewrite, one application of the rule, one exponential.
The general result is $\lim_{x\to\infty}\left(1+\tfrac{k}{x}\right)^{x}=e^{k}$, worth recognising on sight: it runs every continuous-growth model.
A difference of two things that both blow up
Neither term has a limit, so the difference has to be assembled into one object first.
the shape alone does not license the rule; the bottom derivative has to be nonzero on a punctured neighbourhood of the target, and on $0<x<\tfrac{\pi}{2}$ both terms are positive, so it is. After the first application the new bottom is $\sin x+x\cos x$, whose derivative $2\cos x-x\sin x$ falls from $2$ but stays above $0.23$ on $0<x<1$, so the second application is licensed on that smaller window by the same reading. Neither window may be widened by eye: $2\cos x-x\sin x$ changes sign at $x\approx 1.077$, and $\sin x+x\cos x$ turns negative past $x\approx 2.03$. A neighbourhood of the target is all the rule ever asks for, and a window that quietly includes a sign change is the standard way this line goes wrong
Apply the rule, checking the shape between applications
second application; the bottom is now $2$ at $x=0$, so the shape is settled
Answer $$\boxed{L=0}$$
Check
Numerical check at $x=0.1$: $\dfrac{1}{0.1}-\dfrac{1}{\sin 0.1}=-0.016686$, shrinking towards $0$. The limit is $0$ but the approach is from below.
Two infinite quantities can differ by anything. Combining them into one fraction is not cosmetic; it is what creates something the rule can act on.
How fast an angle freezes at a right angle
The angle $\arctan\left(\sqrt3\,x\right)$ settles at $\pi/2$ and never reaches it. What is left over is the interesting quantity, and multiplying it by $x$ measures how fast it disappears. Two separate things have to be said before any derivative: why the outer limit is still $\pi/2$ when the inside is $\sqrt3\,x$ rather than $x$, and which indeterminate form the product is.
a limit passes through a continuous outer function, so $\arctan u\to\pi/2$ whenever $u\to+\infty$; the constant $\sqrt3$ changes the speed of the inside and nothing else, and no derivative is involved in reading this off
the bracket is the difference of two things that agree in the limit, so it shrinks to zero while the factor in front grows; neither factor decides the product, which is what indeterminate means
dividing by $1/x$ is multiplying by $x$. The bracket goes on top because its derivative is one clean fraction; putting the bracket underneath instead would leave a quotient inside a quotient at the next step
chain rule with inner derivative $\sqrt3$, and $\left(\sqrt3\,x\right)^{2}=3x^{2}$; the constant $\pi/2$ differentiates away, which is why subtracting it cost nothing
dividing by a fraction is multiplying by its reciprocal; the single minus sign in the answer comes from $g'(x)=-1/x^{2}$ and from nowhere else, which is the sign check to make at the end
An independent route with no rule at all. For $u>0$ the two angles $\arctan u$ and $\arctan(1/u)$ add to $\pi/2$, so the bracket is exactly $-\arctan\left(1/(\sqrt3\,x)\right)$ and the product is $-x\arctan\left(1/(\sqrt3\,x)\right)$. With $t=1/(\sqrt3\,x)\to 0$ this is $-\dfrac{1}{\sqrt3}\cdot\dfrac{\arctan t}{t}$, and $\arctan t/t\to 1$ because that quotient is the difference quotient of $\arctan$ at $0$, whose value is $1/(1+0)=1$. Numbers: at $x=100$ the product is $-0.57734$ against $-\sqrt3/3=-0.57735$.
One composition limit, one rewrite, one application of the rule.
Two habits leave this example. An arctangent whose inside runs to infinity has a constant limit, so while you are only reading that limit off, read it off the inner behaviour instead of spending a derivative on it. That is not a ban on differentiating it: once the product has been turned into a quotient, the same arctangent is differentiated by the chain rule like anything else, as the third subgoal does. And when a bracket collapses while a factor blows up, the bracket is the one to put on top: differentiate the piece with the tidy derivative.
Checkpoint
§12.7 — choosing the first move for a power form●●●○○
The exponent contains the variable, so this is not a quotient and the rule cannot touch it as written.
Given
$$\displaystyle L=\lim_{x\to 0^{+}}(\sin x)^{x}$$
as $x\to 0^{+}$ the base tends to $0$ and the exponent tends to $0$
Find
(a) What is the correct first move?
Hint 1/4
Ask what shape this is first: base and exponent both head to zero, so it is a power form, not a quotient.
Hint 2/4
Power forms are handled by taking the logarithm: if $L=\lim u^{v}$ then $\ln L=\lim v\ln u$, provided the limit on the right exists.
Hint 3/4
Here $u=\sin x$ and $v=x$, so the logarithm gives $\ln L=\lim_{x\to 0^{+}}x\ln(\sin x)$, which is $0\cdot(-\infty)$ and needs one more rewrite before the rule is allowed.
Hint 4/4
Take logarithms first; the answer, after the dust settles, is $L=1$.
Show solutionTake the logarithm
$$\ln L=\lim_{x\to 0^{+}}x\ln(\sin x)$$
logarithm of a power brings the exponent down
$$=\lim_{x\to 0^{+}}\frac{\ln(\sin x)}{1/x}$$
the product $0\cdot(-\infty)$ becomes $-\infty/\infty$
Numerical check at $x=0.01$: $\sin(0.01)=0.0099998$, and $(0.0099998)^{0.01}=e^{0.01\ln 0.0099998}=e^{-0.04605}=0.955$, moving towards $1$ as slowly as a logarithm does.
⚠ Reporting the logarithm as the answer
The hard work ends when $\ln L$ is found, and the last line feels like bookkeeping.
wrong$$\ln L=3\ \Longrightarrow\ L=3$$
right$$\ln L=3\ \Longrightarrow\ L=e^{3}$$
⚠ Splitting an infinity minus infinity into two limits
The limit laws are usually applied without checking that each piece has a limit.
A question gives you $f$, one value of $f$ and one value of $f'$, and asks about $f^{-1}$.
Check it has an inverse
Show $f'$ keeps one sign on the interval, so $f$ is strictly monotone and one to one.
Walk the output back to its input
Find the $a$ with $f(a)=b$; try small integers. A strictly monotone $f$ has at most one, so a lucky guess is a proof.
Differentiate the original, not the inverse
Compute $f'(a)$ and confirm it is not zero.
Reciprocate
Report $\left(f^{-1}\right)'(b)=1/f'(a)$, and sanity-check the sign against the direction $f$ runs.
Where it goes wrong
Evaluating $f'$ at $b$ instead of at $a$.
Skipping the monotonicity check, so the inverse does not exist and the answer is about nothing.
Meeting $f'(a)=0$ and dividing anyway; that case has a vertical tangent and no derivative.
Naming the form and choosing the move
Substitution produced a shape rather than a number.
Write the hypotheses down before the first derivative
Two lines, and neither is decoration: the shape at the target, and that the bottom derivative is nonzero near it, not at it. A denominator such as $x^{4}$ or $\left(x+\tfrac{\pi}{4}\right)^{3}$ passes, because it vanishes only at the target itself. A condition that is never checked in writing is a condition the marker cannot see you knew.
Substitute and name
Write down what each part tends to. The name of the shape is the whole decision.
If it is $0/0$ or $\infty/\infty$, differentiate top and bottom
Separately, never with the quotient rule.
Otherwise rewrite first
Product: one factor downstairs. Difference: one fraction. Power: logarithm.
Re-check after every application
The shape can stop being indeterminate at any point, and that is the signal to stop and substitute.
Undo what you did
If a logarithm was taken, the answer is $e$ to the limit you just computed.
Where it goes wrong
Applying the rule to a shape that was never indeterminate; it returns a wrong number with no warning.
Differentiating a third time out of momentum, after the shape has already resolved.
Reporting $\ln L$ as $L$.
Ranking functions by order of growth
A box of three expressions with $1$, $2$, $3$ to be written beside them, or any comparison at infinity in which nothing is a plain polynomial and nothing is a plain $e^{x}$. The definition in force is that $f$ grows faster than $g$ when $g/f\to 0$.
Turn every entry into one exponential
Rewrite each expression as $e^{v\ln u}$ using $u^{v}=e^{v\ln u}$, and a quotient of two of them as $e^{P-Q}$. From here on only the exponents matter, because the exponential is increasing and continuous.
Name the inner variable
Almost every entry of this kind is built out of $\ln x$. Put $X=\ln x$, so that $\ln\ln x=\ln X$ and $X\to\infty$, and write each exponent as a combination of $X$, $\ln X$ and constants. An entry such as $3^{\ln x}$ is the same move in the other direction: $3^{\ln x}=x^{\ln 3}$, an ordinary power.
Compare the exponents by the hierarchy
$\ln X$ loses to every positive power of $X$, and $X^{p}$ loses to $X^{q}$ whenever $q>p$. Coefficients and constant bases decide nothing: $2024^{(\ln x)^{0.9}}$ is fast because of the $0.9$, not because of the $2024$. Order the exponents and that is the order of the functions.
Break a tie by subtracting
If two exponents carry the same leading power, subtract them and look at the difference alone: the ratio of the two functions is $e^{\text{difference}}$, so it goes to $0$, to $\infty$, or to a nonzero constant, and that last case is a genuine tie in growth rate.
Write the digits back in the printed order
$1$ beside the slowest and $3$ beside the fastest, placed against the entries as they are printed, not in the order you happened to compare them.
Where it goes wrong
Comparing the expressions themselves instead of their exponents, so that $(\ln x)^{100}$ looks bigger than $x^{0.01}$.
Letting a large constant or a large coefficient decide the ranking.
Judging by the value at $x=10$; every crossing in these tables happens far to the right.
Getting the comparison right and then writing the digits against the wrong entries.
Deciding in advance whether the rule will finish
A printed list of limits with the instruction to circle the ones l'Hopital's rule cannot settle, and any limit where you are about to differentiate for the third time. Wrong circles are subtracted, so the decision has to be made before the work, not after it.
Read the shape at the target
Substitute and write down what each part does. Anything that is not $0/0$ or $\infty/\infty$ is outside the rule: a bounded numerator over $\infty$, a nonzero number over $0$, or a convergent definite integral such as $\int_{0}^{x}dt/(1+t^{2})$ over $x$. These usually have an immediate value.
Apply it once and compare the two quotients
If the new quotient is genuinely simpler, a power lowered or a root cleared, the rule terminates. Say in one sentence how many further applications finish and why, instead of grinding them out; that sentence is what a key pays for.
Count the applications before you start
When the top vanishes to order $n$ at the target and the bottom is the matching power, each application removes one power below and one term above, so exactly $n$ applications finish and the number can be announced in the first line. A third or a fourth derivative is not in itself the warning sign; the warning sign is a quotient that stops getting simpler, which is the next step.
Detect the cycle
If the quotient of derivatives has the shape you started from, up to constants, further applications never simplify. $\sqrt{x^{2}+1}/x$ returns $x/\sqrt{x^{2}+1}$, and $(1+x^{3})^{1/3}/x$ gets worse with every step. The rule is legal here and useless; divide by the dominant term instead.
Detect the silence
If differentiating creates an oscillation with no limit, such as $\cos(1/x)$ or $\sin(1/x)$, then $\lim f'/g'$ does not exist and the rule reports nothing at all. Silence is not a verdict: the original limit may still exist, and it has to be found from the original expression.
Name the replacement tool
Divide by the dominant term, multiply by the conjugate, quote a standard limit such as $\sin x/x\to1$ or $(1-\cos x)/x^{2}\to\tfrac12$, or split off a bounded factor and squeeze. Write the value and the tool that produced it on one line.
Where it goes wrong
Treating "the rule applies" and "the rule helps" as the same thing; the papers separate them and pay for the distinction.
Reporting that the original limit does not exist because $\lim f'/g'$ did not exist.
Circling more items than the question asks for, on a marking table that subtracts for wrong marks.
Differentiating an expression containing $\sin(1/x)$ and trusting the mess that comes back.
Turning a growth limit into a convergence verdict
A multi-part question whose early part asks you to show that some quotient tends to $0$ and whose later part says combine the previous parts to show that an integral converges. The comparison test itself is stated with the improper integrals; what this box supplies is the majorant and the threshold that make it usable, which is the part this week owns.
Make the exponent linear first
A root or a logarithm inside an exponent is what makes the rule endless. Substitute $t=\sqrt{x}$ or $y=\ln x$ so the quotient becomes $t^{n}/e^{t}$, then say once that each application lowers the power by one and reproduces $e^{t}$, so $n$ applications end at $n!/e^{t}\to 0$.
Convert the limit into an inequality with a threshold
From $f/g\to 0$ take one tolerance, say $\varepsilon=1$: there is an $X$ with $0\le f(x)\le g(x)$ for all $x\ge X$. Write that $X$ down. The bound is usually false near the left end, and a key in the pool takes marks off for claiming it there.
Split the integral at the threshold
$\int_{a}^{\infty}=\int_{a}^{X}+\int_{X}^{\infty}$. The first piece is a continuous function on a closed bounded interval, hence a finite number, and only the tail needs a verdict.
Name the test and the comparator out loud
Write the words direct comparison test, and write a complete sentence saying that the comparator converges, with its integral sign and its limits of integration. A key in the pool gives separate marks for the name and for that sentence, and nothing for a scribble.
Write a separate conclusion
One final sentence: the integral converges because the comparator converges and because of the bound past $X$. Pointing at a later part where the integral is evaluated earns nothing, since combining the earlier parts is the question.
Where it goes wrong
Claiming the inequality on the whole interval instead of past a threshold.
Grinding out ten applications of the rule instead of the one sentence that says why they terminate.
Writing "by comparison" without naming the test or the function compared with.
Answering with a number; a comparison delivers a verdict, never a value.
Derivative of x to the power pi
The variable is in the base and the constant is upstairs.
Given
$y=x^{\pi}$, $x>0$
Find
$y'$
SolutionRecognise the position of the variable
$$y=x^{\pi}$$
constant exponent, so this is the power rule, valid for every real exponent when $x>0$
Apply the power rule
$$y'=\pi x^{\pi-1}$$
bring the exponent down and reduce it by one
Answer $$\boxed{y'=\pi x^{\pi-1}}$$
Check
Check through the definition: $x^{\pi}=e^{\pi\ln x}$, so $y'=e^{\pi\ln x}\cdot\dfrac{\pi}{x}=\pi x^{\pi-1}$. The power rule for irrational exponents follows; it is not an extra assumption.
Derivative of pi to the power x
The same two symbols, swapped.
Given
$y=\pi^{x}$
Find
$y'$
SolutionRecognise the position of the variable
$$y=\pi^{x}$$
constant base, variable exponent, so this is the general exponential rule
Apply the exponential rule
$$y'=\pi^{x}\ln\pi$$
the function survives and picks up the factor $\ln$ of the base
Answer $$\boxed{y'=\pi^{x}\ln\pi}$$
Check
Check at $x=0$: the tangent to any $a^{x}$ at $(0,1)$ has slope $\ln a$, and the formula gives $\ln\pi\approx 1.145$ — steeper than $e^{x}$, as a bigger base should be.
Same two symbols, two different rules, and the answers do not even have the same shape: a power of $x$ against an exponential.
How to tell them apart
Point at the $x$. In the base means power rule; in the exponent means multiply by $\ln$ of the base; in both means take logarithms first.
Scaffolding comes off
The common skeleton
Substitute the limit point and write down the shape you get.
If the shape is $0/0$ or $\infty/\infty$, differentiate top and bottom separately.
If it is a product, a difference or a power, rewrite it into one of those two shapes first.
After every application, substitute again and re-name the shape.
Undo any transformation you made, and check the size of the answer against a value near the limit point.
1 · fully worked
A product where one factor vanishes and the other blows up
The full solution, every reason written out.
Given
$$\displaystyle L=\lim_{x\to 0^{+}}x^{2}\ln x$$
Find
$L$
SolutionName the shape
$$x^{2}\to 0,\qquad \ln x\to-\infty$$
the shape is $0\cdot(-\infty)$, which the rule does not accept
Move one factor downstairs
$$x^{2}\ln x=\frac{\ln x}{x^{-2}}$$
the factor that blows up stays on top; putting $x^{2}$ downstairs as $x^{-2}$ is what turns a product into a quotient
simplify first, substitute second; the shape is no longer indeterminate
Answer $$\boxed{L=0}$$
Check
Numerical check at $x=0.001$: $x^{2}\ln x=10^{-6}\times(-6.908)=-6.9\times 10^{-6}$, already within seven millionths of zero. The power beats the logarithm, which is the general lesson.
Which factor goes downstairs is a choice; sending $\ln x$ down instead produces a mess that never resolves.
2 · you write the reasoning
The same skeleton on a lighter problem: $$\displaystyle\lim_{x\to 0^{+}}x\ln x$$. Steps given, reasons not. Say why each line is legal before opening the model answer.
Substituting gives $0\cdot(-\infty)$, so no rule applies yet.
reasoning
Substitution is always the first move, and here it produces a shape rather than a number: $0\cdot(-\infty)$ is not something the rule accepts.
Rewrite as a quotient: $$\displaystyle\lim_{x\to 0^{+}}\frac{\ln x}{1/x}$$.
reasoning
Dividing by $1/x$ is multiplying by $x$, so nothing changed except the arrangement — and the arrangement is now a quotient.
The shape is now $-\infty/\infty$, so differentiate top and bottom.
reasoning
Both parts now run to infinity in size, the second shape the rule accepts, so differentiating top and bottom separately is legal.
Simplify and substitute: the answer is $0$.
reasoning
The quotient $\dfrac{1/x}{-1/x^{2}}$ simplifies to $-x$ before any limit is taken; simplifying first is what keeps the last step trivial.
3 · find the buried error
A worked solution to $$\displaystyle L=\lim_{x\to 0^{+}}(1+2x)^{1/x}$$ is written below. It reaches an answer, and the answer is wrong. Two separate errors are buried in it.
Step 1. As $x\to 0^{+}$ the base tends to $1$ and the exponent $1/x$ tends to $\infty$, so the shape is $1^{\infty}$ and we take logarithms.
Step 2. Written as a quotient this is $\dfrac{\ln(1+2x)}{x}$, and substituting gives $\dfrac{0}{0}$, so the rule applies.
Step 3. Differentiating, $\dfrac{d}{dx}\ln(1+2x)=\dfrac{1}{1+2x}$ and $\dfrac{d}{dx}x=1$, so $\ln L=\lim_{x\to 0^{+}}\dfrac{1}{1+2x}=1$.
Step 4. Therefore $L=1$.
the two buried errors (2)
⚠ step 3
The chain rule was dropped inside the logarithm: $\dfrac{d}{dx}\ln(1+2x)=\dfrac{2}{1+2x}$, not $\dfrac{1}{1+2x}$.
The outer derivative of $\ln$ is the visible one and the inner factor is a single digit, so it disappears unnoticed — the same slip costs the same mark in every logarithmic differentiation question.
right
With the factor restored, $\ln L=\lim_{x\to 0^{+}}\dfrac{2}{1+2x}=2$.
⚠ step 4
The logarithm was never undone: even with the value found in step 3, the answer would be $L=e^{1}$, not $L=1$.
After the hard part is over, the final exponential feels like bookkeeping rather than mathematics, and $\ln L$ is quietly read as $L$.
right
The correct last line is $L=e^{2}\approx 7.389$.
4 · the bare problem
§12.7 — a bare power form, no scaffolding●●●○○
Nothing is set up for you here. Name the shape, choose the move, and finish the job including the last line.
Given
$$\displaystyle L=\lim_{x\to\infty}x^{1/x}$$
Find
(a) Evaluate $L$, showing the shape at each stage.
Hint 1/4
Look at base and exponent separately as $x$ grows: one runs to infinity, the other to zero. That combination is a power form.
Hint 2/4
For a power form, take the logarithm: $\ln L=\lim v\ln u$ with $u=x$ and $v=1/x$. Undo it at the end.
Hint 3/4
Here $\ln L=\lim_{x\to\infty}\dfrac{\ln x}{x}$, which is $\infty/\infty$ and ready for the rule; the data again: $u=x$, $v=1/x$.
Hint 4/4
$\ln L=0$, so $L=e^{0}=1$.
Show solutionName the shape
$$x\to\infty,\quad \frac{1}{x}\to 0$$
the shape is $\infty^{0}$: a power form, not a quotient
the exponent comes down; the result is already a quotient of shape $\infty/\infty$
Apply the rule and undo the logarithm
$$\ln L=\lim_{x\to\infty}\frac{1/x}{1}=0$$
differentiate top and bottom
$$L=e^{0}=1$$
exponentiate
Answer $$\boxed{L=1}$$
Check
Numerical check at $x=1000$: $1000^{0.001}=e^{0.0069078}=1.00693$, and at $x=10^{6}$ it is $1.0000138$. The approach is slow, as anything driven by a logarithm is.
$\infty^{0}$ came out $1$ here, but the shape did not decide that — the rate did. Another pair with the same shape can give any positive answer.
Full exam-style question
Three parts in the shape a final examination asks themexam format
One question touching all three families. Nothing is harder than the worked examples; the difficulty is that the three parts do not announce which method they want.
Given
$h(x)=\arctan\left(x^{2}\right)+5^{x}$
$f(x)=x+\arcsin x$ on $(-1,1)$, which is strictly increasing there
Part (a) at $x=0$: the formula gives $\ln 5\approx 1.609$, and $h$ near zero is $5^{x}$ plus something quadratically small. Part (c) at $x=0.05$: $(\cos 0.05)^{400}=0.60640$ against $e^{-1/2}=0.60653$.
Three parts, three different first moves; the recognition is the examined skill.
Notice what part (b) did not need: a formula for $f^{-1}$. A strictly increasing function plus one convenient value is always asking for the reciprocal-slope theorem.
Practice
A · concept 3 questions
1§12.1 — which slot the number goes into●●○○○
The data below is everything a question of this kind gives you. The claim underneath uses all of it and still gets the answer wrong.
Given
$f$ is one to one and differentiable on $\mathbb{R}$
$f(2)=5$
$f'(2)=4$
Find
(a) True or false: it follows that $\left(f^{-1}\right)'(2)=\tfrac14$.
(b) Whatever you answered, write down the one statement about $f^{-1}$ that the data does support.
Hint 1/4
Draw the arrow: $f$ sends $2$ to $5$. Now ask which of those two numbers the inverse takes as its input.
Hint 2/4
The rule is $\left(f^{-1}\right)'(b)=1/f'\!\left(f^{-1}(b)\right)$, so the number written inside $\left(f^{-1}\right)'$ is an output of $f$.
Hint 3/4
The data again: $f(2)=5$ and $f'(2)=4$, so the reciprocal $\tfrac14$ belongs at the output $5$.
Hint 4/4
False; the supported statement is $\left(f^{-1}\right)'(5)=\tfrac14$.
Consistency check with slopes: $f$ climbs $4$ units of output per unit of input at $x=2$, so the mirrored graph climbs a quarter of a unit near the mirrored point.
2§12.7 — what a shape does and does not decide●●○○○
Two limits with the same shape are placed side by side so that the claim can be tested rather than argued about.
Check the first term for evenness: $\arctan\left(3x^{2}\right)$ is even, so its derivative must be odd and must vanish at $0$. The expression $6x/(1+9x^{4})$ does both.
2§12.3 — arcsine of a square root●●●○○
A composite whose inner function has its own domain restriction, so the answer comes with an interval attached.
Given
$y=\arcsin\left(\sqrt{x}\right)$
Find
(a) Differentiate and simplify.
(b) State the interval on which your answer is valid.
Hint 1/4
Identify the inner function and note where it is allowed to live before differentiating.
Hint 2/4
$(\arcsin u)'=\dfrac{u'}{\sqrt{1-u^{2}}}$ with $u=\sqrt{x}$, whose derivative is $\dfrac{1}{2\sqrt{x}}$.
Hint 3/4
Substituting: $u^{2}=x$, so the root becomes $\sqrt{1-x}$, and the product is $\dfrac{1}{\sqrt{1-x}}\cdot\dfrac{1}{2\sqrt{x}}$.
Hint 4/4
$y'=\dfrac{1}{2\sqrt{x(1-x)}}$ on $0<x<1$.
Show solutionChain rule with the inner square root
Symmetry check that does not repeat the computation: $\arcsin\sqrt{x}+\arcsin\sqrt{1-x}=\tfrac{\pi}{2}$, so differentiating forces $y'(x)=y'(1-x)$ — and the answer is unchanged when $x$ and $1-x$ swap.
Whenever an inner function has a restricted domain, the interval is part of the answer, not decoration.
3§12.2 — variable in the base and in the exponent●●●○○
Neither the power rule nor the exponential rule applies, because the variable is in both positions at once. There is exactly one move.
Given
$y=x^{\tan x}$ for $x>0$ and $\cos x\neq 0$
Find
(a) Find $y'$.
(b) Evaluate $y'$ at $x=1$, to three decimals.
Hint 1/4
Check both positions of the variable before choosing a rule; when it sits in both, no memorised derivative applies.
Hint 2/4
Take logarithms of both sides and differentiate implicitly: $\ln y=v\ln u$ gives $\dfrac{y'}{y}=(v\ln u)'$.
Hint 3/4
Here $\ln y=\tan x\ln x$, whose derivative by the product rule is $\sec^{2}x\ln x+\dfrac{\tan x}{x}$, and $y$ itself is $x^{\tan x}$.
Hint 4/4
$y'=x^{\tan x}\left(\sec^{2}x\ln x+\dfrac{\tan x}{x}\right)$, which at $x=1$ is $\tan 1\approx 1.557$.
Show solutionTake logarithms to separate the two positions
$$\ln y=\tan x\ln x$$
the logarithm turns a variable exponent into a factor, which is the only reason this problem is doable
Numerical check at $x=1$ with a symmetric difference of step $0.01$: $\dfrac{1.01^{\tan 1.01}-0.99^{\tan 0.99}}{0.02}=1.5583$, against $\tan 1=1.5574$.
This is the third row of the table in the general-base block, and it is always this move.
4§12.6 — two applications with a check between them●●●○○
A quotient that stays indeterminate after the first differentiation, so the shape has to be read twice.
(a) Evaluate $L$, naming the shape before each application.
Hint 1/4
Substitute first and write down the shape; that decides whether you may start at all.
Hint 2/4
The rule differentiates top and bottom separately, and may be repeated while the shape stays indeterminate.
Hint 3/4
Here $\tfrac{1-\cos 0}{0}=\tfrac{0}{0}$, and one application leaves $\dfrac{\sin x}{2x}$, which is $\tfrac00$ again.
Hint 4/4
The second application gives $\dfrac{\cos x}{2}\to\tfrac12$.
Show solutionShape check, then first application
$$\frac{1-\cos 0}{0^{2}}=\frac{0}{0}$$
indeterminate, so the rule may start
$$L=\lim_{x\to 0}\frac{\sin x}{2x}$$
top and bottom differentiated separately, not as a quotient
Shape check, then second application
$$\frac{\sin 0}{0}=\frac{0}{0}$$
still indeterminate, so a second round is legal
$$L=\lim_{x\to 0}\frac{\cos x}{2}=\frac12$$
substitution works now, so stop
Answer $$\boxed{L=\tfrac12}$$
Check
Numerical check at $x=0.05$: $\dfrac{1-\cos 0.05}{0.0025}=\dfrac{0.00124974}{0.0025}=0.49990$.
Stopping is a decision too. Once the shape is a number the rule is no longer allowed, and a third differentiation would replace a correct answer with a meaningless one.
5§12.7 — a difference of two quantities that both blow up●●●●○
Neither term has a limit on its own, so the limit laws do not apply until the two are made into one object.
Numerical check: $f(1.02)=1.061208+3.06+2=6.121208$, so near the output $6$ the inverse gains $0.02$ of input for $0.121208$ of output, a rate of $0.165$ against the exact $0.1667$.
The coincidence $f(1)=6$ and $f'(1)=6$ is bait: one is an output, the other a slope, and only the second gets reciprocated.
2§12.5 — the hyperbolic tangent, end to end●●●●○
Three parts on one function, in the order an examination builds them: a structural fact, an inverse derivative by two routes, then a limit needing the rule twice.
Given
$\tanh x=\dfrac{\sinh x}{\cosh x}$
$\tanh^{-1}x=\tfrac12\ln\dfrac{1+x}{1-x}$ for $\vert x\vert<1$
$\cosh^{2}x-\sinh^{2}x=1$
Find
(a) Show that $\tanh$ is strictly increasing on $\mathbb{R}$.
(b) Find $\left(\tanh^{-1}\right)'(x)$ twice: once from the reciprocal-slope theorem and once by differentiating the logarithmic form.
Part (a) is about a sign, part (b) about doing one job two ways, part (c) about naming a shape before touching it.
Hint 2/4
Use $(\tanh x)'=\operatorname{sech}^{2}x$ for (a); $\left(f^{-1}\right)'=1/f'\!\left(f^{-1}\right)$ for (b); and l'Hôpital's rule, repeatedly and with a check each time, for (c).
Hint 3/4
In (b) the theorem needs $\operatorname{sech}^{2}y$ where $\tanh y=x$, and $\operatorname{sech}^{2}y=1-\tanh^{2}y=1-x^{2}$; the logarithmic form is $\tfrac12\left[\ln(1+x)-\ln(1-x)\right]$.
Hint 4/4
The answers are: $\operatorname{sech}^{2}x>0$; $\dfrac{1}{1-x^{2}}$ by both routes; and $-\tfrac13$.
Two independent checks. Part (b): the two routes were computed by different methods and agree. Part (c) numerically at $x=0.1$: $\dfrac{\tanh 0.1-0.1}{0.001}=-0.3320$, against $-\tfrac13=-0.3333$.
The answer $-\tfrac13$ says that $\tanh x\approx x-\tfrac{x^{3}}{3}$ near zero, which is why $\tanh$ falls away from the line $y=x$ so quickly.
3§12.7 — a power form under examination conditions●●●○○
Exponent and base both move, and the answer is one of four numbers, each matching a different slip in the same method.
Given
$$\displaystyle L=\lim_{x\to 0^{+}}(1+3x)^{2/x}$$
Find
(a) Which value is $L$?
Hint 1/4
Name the shape first: as $x\to 0^{+}$ the base tends to $1$ and the exponent to $\infty$.
Hint 2/4
For a power form, $\ln L=\lim v\ln u$, and the answer is $e$ to whatever that limit is.
Hint 3/4
Here $\ln L=\lim_{x\to 0^{+}}\dfrac{2\ln(1+3x)}{x}$, a $\tfrac{0}{0}$ quotient; the data again: base $1+3x$, exponent $2/x$.
Independent check of the slope: the curve climbs from $0.78540$ at $x=1$ to $0.80520$ at $x=1.02$, a measured average of $0.99$, consistent with a slope of exactly $1$ at the left end.
Linear approximation and the new derivative table are one question here, which is how the material actually gets examined.
2§12.3 — an angle changing in time●●●○○
A camera on a straight road tracks a car. Distances in metres, time in seconds, angle measured from the perpendicular to the road.
Given
the camera stands $50$ m from the road, at the closest point
$x$ is the car's distance along the road from that closest point
$\tan\theta=\dfrac{x}{50}$
$\dfrac{dx}{dt}=20$ m/s
at the instant asked, $x=50$ m
Find
(a) How fast is $\theta$ changing at that instant, in radians per second?
Hint 1/4
Two quantities change in time and one equation ties them together, so the first step is to write $\theta$ as a function of $x$.
Hint 2/4
$\theta=\arctan\left(\dfrac{x}{50}\right)$, and the chain rule in time gives $\dfrac{d\theta}{dt}=\dfrac{d\theta}{dx}\cdot\dfrac{dx}{dt}$.
Hint 3/4
With $\dfrac{d\theta}{dx}=\dfrac{1}{1+\left(x/50\right)^{2}}\cdot\dfrac{1}{50}$ and the data $x=50$, $\dfrac{dx}{dt}=20$: the bracket is $1+1=2$.
Size check without the formula: at $x=50$ the car is at $45$ degrees, $50\sqrt2\approx 70.7$ m away, and its velocity across the line of sight is $20\cos 45^{\circ}\approx 14.1$ m/s; $14.1/70.7=0.2$ rad/s.
Inverse trigonometric derivatives are where related rates and this section meet, and the meeting is common in examinations.
3§12.3 — a derivative of an integral, unannounced●●●●○
The variable appears as a limit of integration rather than inside the integrand, which changes which theorem opens the problem.
Size check: on $[0,\pi/4]$ the integrand $e^{t^{2}}$ lies between $1$ and $1.85$, so a derivative just under $1$ at $x=1$ is the right order, and positive as an increasing $G$ requires.
No antiderivative of $e^{t^{2}}$ exists in elementary terms, which is the reason the question is phrased this way.
Shaped like the real papers 4 questions
1§12.2 — ordering by growth, the way the final asks it●●●●○
Two boxes, six entries, one digit beside each entry. The paper hands you the definition of grows faster and wants the ranking, not the limits themselves, and it pays per box. Nothing here is a polynomial and nothing here is $e^{x}$, so the usual size intuition is no help. Budget six minutes.
Given
Definition in force: $f$ grows faster than $g$ means $g/f\to 0$ as $x\to\infty$.
Every entry is positive for large $x$, so ratios may be compared through their logarithms.
Find
(a) In each box write $1$, $2$ or $3$ beside each entry, $1$ for the slowest and $3$ for the fastest.
(b) In one sentence, say what makes $\dfrac{x^{0.01}}{(\ln x)^{50}}$ beat $5^{\sqrt{\ln x}}$, and name the rule that settles the comparison.
Hint 1/4
Every entry is built out of one quantity, and that quantity is itself running to infinity. Ask what the natural variable of this problem is before you compare anything.
Hint 2/4
For $u>0$, $u^{v}=e^{v\ln u}$. Once each entry is $e^{(\text{exponent})}$, one entry beats another exactly when the difference of the exponents runs to $+\infty$.
Hint 3/4
Put $X=\ln x$, so that $\ln\ln x=\ln X$ and $X\to\infty$. Box A becomes $e^{X/\ln X}$, $e^{\sqrt{X}\ln X}$, $e^{(\ln 2)X^{0.7}}$; Box B becomes $e^{(\ln 5)\sqrt{X}}$, $e^{0.01X-50\ln X}$, $e^{100\ln X}$.
Hint 4/4
Compare the exponents as powers of $X$: $\ln X$ loses to every positive power of $X$, and between two powers the larger exponent wins. Reading the entries in the order printed, Box A ranks $3,1,2$ and Box B ranks $2,3,1$.
Show solutionChoose the variable the problem is actually written in
the linear term wins, so the quotient tends to zero, which by the stated definition is what beats means
Check
Box B at $X=\ln x=10^{6}$: the exponents are $100\ln X=1382$, $(\ln 5)\sqrt{X}=1609$ and $0.01X-50\ln X=9310$, which is the claimed order. Box A settles far later. At $X=100$ the exponents of $x^{1/\ln\ln x}$, $(\ln x)^{\sqrt{\ln x}}$ and $2^{(\ln x)^{0.7}}$ are $21.7$, $46.1$ and $17.4$, so at that stage the printed order would read $2,3,1$, which is not the answer. At $X=10^{8}$ the same three exponents are $5.43\times10^{6}$, $1.84\times10^{5}$ and $2.76\times10^{5}$, and now the printed order reads $3,1,2$ as claimed. The ranking is a statement about the limit, not about any particular $x$ you can write down.
2§12.6 — where the rule stops being useful●●●●○
A list, four labels, and a marking table that subtracts for wrong marks: this is the shape the final uses to test whether you check the hypothesis before you reach for the rule. Two of the six have perfectly good limits that the rule will never hand you.
Given
Label $\mathrm{W}$: the rule applies and finitely many applications finish the limit.
Label $\mathrm{F}$: the expression as written is neither $0/0$ nor $\infty/\infty$, so the rule does not apply to it.
Label $\mathrm{N}$: the rule applies, but the quotient of derivatives has no limit, so the rule returns no information.
Label $\mathrm{C}$: the rule applies, but repeated application never simplifies the expression.
(b) Items $2$ and $5$ both have limits. Give both values, one line each, naming the tool that produced it.
Hint 1/4
Before differentiating anything, read every expression at its target point and write down the shape you actually see there. Two of the six are decided by that reading alone.
Hint 2/4
The rule needs $0/0$ or $\infty/\infty$ at the target, and it reports back only when $\lim f'/g'$ exists or is infinite. If that limit fails to exist, the rule says nothing at all; and if $f'/g'$ keeps the shape of $f/g$ forever, it will keep saying nothing.
Hint 3/4
Differentiate items $2$ and $4$ once each: item $2$ gives $\dfrac{1-\sin x}{1+\sin x}$, and item $4$ gives $\dfrac{x}{\sqrt{x^{2}+1}}$, whose own quotient of derivatives is $\dfrac{\sqrt{x^{2}+1}}{x}$ again.
Hint 4/4
Item $1$ is $\mathrm{W}$, items $2$ and $5$ are $\mathrm{N}$, items $3$ and $6$ are $\mathrm{F}$, item $4$ is $\mathrm{C}$. For the two values, divide through by $x$ in item $2$, and split off the bounded factor in item $5$.
the quotient of derivatives is a bounded oscillation that keeps returning to $0$ and to $\infty$ along different sequences of $x$
$$\lim_{x\to\infty}\frac{1-\sin x}{1+\sin x}\ \text{does not exist}$$
the hypothesis that this limit exists fails, so the rule licenses no conclusion at all; the label is $\mathrm{N}$, and note that $1+\sin x$ also vanishes arbitrarily far out, which breaks a second hypothesis
dividing top and bottom by $x$ and using $\sin x/x\to 1$; the surviving $\sin(1/x)$ has no limit, so the rule returns no information and the label is $\mathrm{N}$
one label each, with three different reasons for the five items the rule does not settle
Check
Item $5$ at $x=0.01$: $x^{3}\cos(1/x)=10^{-6}\times 0.8623=8.62\times10^{-7}$ and $1-\cos x=5.00\times10^{-5}$, so the quotient is $0.0172$, already heading for $0$; meanwhile the quotient of derivatives at that same point evaluates to $-0.48$, and since it is $3x\cos(1/x)+\sin(1/x)$ up to a factor tending to $1$, it will still be swinging across $[-1,1]$ however far in you go. Item $6$ rationalises to $x/(\sqrt{x^{2}+x}+x)\to 1/2$, so it too has a limit the rule was never entitled to look for.
3§12.6 — a limit that exists in order to license a convergence claim●●●●●
The limit is not the answer here, it is the permission slip. Marks sit on the argument that finitely many applications finish, on naming the test out loud, and on stating a threshold honestly instead of claiming a bound everywhere. Fifteen minutes, and the last part asks for a setup only.
$x^{3}e^{-\sqrt{x}}$ is continuous and positive on $[1,\infty)$.
$$\displaystyle\int_{X}^{\infty}\frac{dx}{x^{2}}$$ converges for every $X>0$.
Find
(a) Show that $$\displaystyle\lim_{x\to\infty}\frac{x^{5}}{e^{\sqrt{x}}}=0$$, making it clear why finitely many applications of the rule reach the answer.
(b) Use part (a) to show that $J$ converges. Name the test, write down the inequality you compare with, and be explicit about where that inequality starts holding.
(c) Write down, but do not carry out, the substitution that turns $J$ into an integral you could finish by parts, and say how many applications of integration by parts it would take.
Hint 1/4
A square root sitting inside an exponent is unpleasant to differentiate over and over, and easy to remove. Decide what variable makes that exponent linear before you commit to any method.
Hint 2/4
After the change of variable the quotient is $t^{n}/e^{t}$ with $t\to\infty$, and each application of the rule leaves the denominator untouched while lowering the power on top by one. For part (b), remember what a limit of $0$ actually promises: for any tolerance you name there is an $X$ beyond which the quantity stays below it.
Hint 3/4
Put $t=\sqrt{x}$, so $x=t^{2}$ and $x^{5}/e^{\sqrt{x}}=t^{10}/e^{t}$. For part (b) the useful algebra is $x^{3}e^{-\sqrt{x}}=\dfrac{x^{5}}{e^{\sqrt{x}}}\cdot\dfrac{1}{x^{2}}$.
Hint 4/4
Ten applications leave $10!/e^{t}\to 0$. Taking the tolerance $1$ in part (a) produces an $X$ with $x^{3}e^{-\sqrt{x}}\le x^{-2}$ for $x\ge X$, and the direct comparison test closes part (b). In part (c), $t=\sqrt{x}$ turns $J$ into $2\int_{1}^{\infty}t^{7}e^{-t}\,dt$, which needs seven applications of parts.
Show solutionMake the exponent linear before differentiating
each application differentiates $e^{t}$ into itself and drops the power on top by exactly one, so the shape stays $\infty/\infty$ and the exponent counts down
the shape is now a constant over something unbounded, which is no longer indeterminate, so the chain ends and the answer is $0$
Turn the limit into a bound, with the threshold written down
$$\exists X\ge 1:\quad \frac{x^{5}}{e^{\sqrt{x}}}\le 1\quad\text{for all }x\ge X$$
this is what part (a) means with the tolerance taken to be $1$; the threshold cannot be dropped, and claiming the bound for all $x\ge 1$ would be false
$x^{3}=t^{6}$ and $dx=2t\,dt$, so the polynomial factor is $t^{7}$
$$7\ \text{applications}$$
each application of parts differentiates the polynomial factor once, lowering its degree by one, and the remaining $\int e^{-t}dt$ is elementary
Check
The threshold in part (b) is not decoration. At $x=1000$ we have $x^{5}=10^{15}$ while $e^{\sqrt{1000}}=e^{31.62}=5.4\times10^{13}$, so the bound $x^{5}\le e^{\sqrt{x}}$ is false there; it turns true near $x\approx 1.3\times10^{3}$ and is then overwhelming, since at $x=10^{4}$ we get $10^{20}$ against $e^{100}=2.7\times10^{43}$. A solution that writes the inequality for all $x\ge 1$ has written something untrue, which is why the threshold carries its own mark.
4§12.7 — power forms settled by exponents rather than by $\ln L$●●●●●
A fixed base with a variable exponent upstairs, a variable base with a variable exponent downstairs, and a constant to solve for. Taking $\ln L$ and differentiating is legal and will cost you the question; the route the quiz rewards is to write the whole quotient as one exponential and read what the single exponent does. Ten minutes.
$\dfrac{\ln\ln x}{\ln x}\to 0$ and $\dfrac{x}{\ln x}\to\infty$ as $x\to\infty$.
Find
(a) Write the quotient as a single $e^{\square}$ and simplify the exponent until no variable base is left anywhere.
(b) Find every $a>0$ for which $L(a)=0$, treating the borderline value on its own.
(c) Name the property of the exponential function that lets you pass from the behaviour of the exponent to the value of $L(a)$.
Hint 1/4
Only the numerator is an exponential with a fixed base; the denominator has a variable base, and neither floor is a power with a fixed exponent. Settle what a quantity like $u^{v}$ even means before trying to compare two of them.
Hint 2/4
For $u>0$, $u^{v}=e^{v\ln u}$, and $e^{p}/e^{q}=e^{p-q}$. So a quotient of two such objects is one exponential, and its size is decided entirely by where that one exponent goes.
Hint 3/4
The denominator's exponent is $\dfrac{x}{\ln x}\bigl(\ln x+\ln\ln x\bigr)=x+\dfrac{x\ln\ln x}{\ln x}$, and the numerator's is $x\ln a$.
Hint 4/4
The exponent is $x\left(\ln a-1-\dfrac{\ln\ln x}{\ln x}\right)$. The bracket tends to $\ln a-1$, so the sign of $\ln a-1$ decides every case except $a=e$, where the bracket is exactly $-\dfrac{\ln\ln x}{\ln x}$ and the exponent is $-\dfrac{x}{\ln x}\ln\ln x\to-\infty$. So $L(a)=0$ precisely for $0<a\le e$.
Show solutionPut both floors into exponential normal form
$$a^{x}=e^{x\ln a}$$
a constant base other than $e$ goes through $e$ and $\ln$, which is the only way the base can be compared with anything
both factors are unbounded for $x>e^{e}$, using the given fact $x/\ln x\to\infty$; a tie in the leading term does not make the limit $1$
$$L(e)=0$$
so the borderline joins the zeros rather than separating them
Name what licenses the last step
$$u(x)\to-\infty\ \Rightarrow\ e^{u(x)}\to 0$$
the exponential is continuous on $\mathbb{R}$, so limits pass through it by the composition rule, and its end behaviour supplies the two infinite cases; this is the property part (c) asks to be named
Take $a=3$, which is above $e$. At $x=e^{100}$ the bracket is $\ln 3-1-\frac{\ln 100}{100}=1.0986-1-0.0461=0.0525>0$, so the exponent is $0.0525\,e^{100}$ and the quotient is enormous, as claimed. At the far smaller $x=e^{10}$ the same bracket is $1.0986-1-0.2303=-0.132$, still negative, so the quotient is tiny there: the answer only settles once $\ln\ln x/\ln x$ has dropped below $\ln a-1$. For $a=e$ at $x=e^{100}$ the exponent is $-\frac{e^{100}}{100}\ln 100=-0.046\,e^{100}$, confirming that the tie still falls to zero.
Mistake ledger (16 entries)
⚠ Evaluating the derivative at the output instead of the input
The number $b$ is the one in the question, so the eye puts it in every slot it fits.
⚠ Buried error found in the scaffolding ladder, step 3
The chain rule was dropped inside the logarithm: $\dfrac{d}{dx}\ln(1+2x)=\dfrac{2}{1+2x}$. The outer derivative is the visible one and the inner factor is a single digit, so it disappears unnoticed.
⚠ Buried error found in the scaffolding ladder, step 4
The logarithm was never undone: with the step-3 value the answer would be $L=e^{1}$, not $L=1$. Once the hard part is over, $\ln L$ is quietly read as $L$.
$f$ one to one and differentiable near $a=f^{-1}(b)$, and $f'(a)\neq 0$. That last line is not decoration: when $f'(a)=0$ the inverse has no derivative at $b$ at all, whatever $f$ is. Suppose it had one; differentiating $f^{-1}\!\left(f(x)\right)=x$ at $x=a$ gives $\left(f^{-1}\right)'(b)\cdot f'(a)=1$, that is $0=1$. The difference quotient says the same thing in pictures: it is the reciprocal of a quotient that tends to $0$ and is never $0$, so it grows without bound and the horizontal tangent of $f$ reflects into a vertical tangent of $f^{-1}$. One counterexample settles one claim; this settles the true or false version about every $f$.
$u$ differentiable. This is the general base rule at $a=e$, where the constant factor $\ln a$ becomes $\ln e=1$ and disappears, so nothing is left but the inner derivative.
Power rule for a real exponent
$$\frac{d}{dx}x^{n}=nx^{n-1}$$
Any real $n$, for $x>0$. The restriction belongs to the derivation through $e^{n\ln x}$, not to the rule: when $n$ is a positive integer the same formula follows from the product and sum rules built earlier in the course and holds for every real $x$, negative values included. So a two sided limit resting on $x^{3}$ or $x^{4}$ may be differentiated from both sides of $0$ without borrowing anything from this box.
Any $A>0$, any $p>0$, any $a>1$, any $n>0$: a power of a logarithm always loses to a power of $x$, and a power of $x$ always loses to an exponential. Each one is finitely many applications of the rule: each application lowers the power on top by one while reproducing the bottom up to a constant factor, and the chain stops as soon as that power reaches zero or goes negative. On the third line, with a whole number $n$, the constants multiply out to exactly $n!$ after $n$ steps. The other two collect different constants — a whole number $A$ on the first line ends at $A!/p^{A}$ over $x^{p}$, and the second gathers $\left(\ln a\right)^{n}$ on the bottom — and a power that is not a whole number lands on no factorial at all; it simply runs out once the exponent turns negative. Use it to rank functions, to settle an $\infty/\infty$ limit without differentiating ten times, and to supply the majorant in a convergence argument; the sentence about why the applications terminate is itself a marked step. A sketch of a function built from $\ln$ and $e^{x}$ is where these limits are spent: the graph-shape section settles each edge of the domain by its one sided limit, and a finite value there is an open end with an entry slope to draw, not an asymptote — the limit is the only part of that checklist the earlier sections could not compute.
Needs $u>0$ and $a>0$; the third holds because the exponential is continuous, with $e^{-\infty}=0$ and $e^{+\infty}=\infty$. Use it whenever a variable base carries a variable exponent, on one floor or on both: push every floor into one exponential, subtract the exponents, and read that single exponent. This is faster than setting $L=\lim u^{v}$, taking $\ln L$ and differentiating, and it is the only route that finishes a ranking table inside the time given.
$x$ in radians; each is the value the rule returns, so quoting it is legitimate and quicker, and each still needs its own $0/0$ reading before use. Use them when one of these blocks appears as a factor of a larger expression that the rule cannot touch as a whole, typically next to a bounded oscillation such as $\sin(1/x)$: quote the standard value for the factor and handle the oscillating factor separately.
$\left|\arctan x\right|<\pi/2$ for every real $x$, so an arctangent inside a limit is a bounded quantity that freezes to $\pm\pi/2$, and that value is read off rather than produced by differentiating. The reading replaces a derivative only while the limit itself is what you want; inside an application of the rule the same arctangent is differentiated normally, $\left(\arctan u\right)'=u'/\left(1+u^{2}\right)$. Use it when an arctangent sits in an exponent, and to see that a numerator such as $\int_{0}^{x}dt/(1+t^{2})$ stays finite, so a quotient with that numerator over $x$ is a finite number over $\infty$, not $\infty/\infty$, and the rule does not apply to it at all. The values survive a composition: $u(x)\to+\infty$ gives $\arctan u(x)\to\pi/2$ and $u(x)\to-\infty$ gives $-\pi/2$, since a limit passes through a continuous outer function, so an inner function running to infinity is read off the same two numbers.
From a limit to an inequality valid beyond a threshold
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0\ \Longrightarrow\ \text{for every }\varepsilon>0\text{ there is an }X\text{ with }0\le f(x)\le\varepsilon\,g(x)\text{ for all }x\ge X$$
$f\ge 0$ and $g>0$ near infinity; if the limit is a number $L>0$ instead of $0$, the same reading gives $\tfrac{L}{2}g(x)\le f(x)\le 2Lg(x)$ past some $X$. The inequality is promised only past $X$ and is usually false before it. Use it whenever a growth limit has to license a bound, a domination claim or the convergence of an integral, and write the $X$ down: stating $x>X$ honestly instead of claiming the bound everywhere carries its own mark.
$a>0$, $a\neq 1$, $u$ differentiable, and $u>0$ for the logarithm. The bare $a^{x}$ form on the card above is almost never the form a paper hands you: the exponent is $2x$, or $x^{2}$, or $\sqrt{x}$, and the two places marks disappear are the missing $u'$ and the missing $\ln a$. Use it whenever a constant base carries anything other than a bare $x$, including inside a tangency or threshold condition where the derivative is the tool rather than the answer.
Every real $x$. Both are the addition formula $\sin(A+B)=\sin A\cos B+\cos A\sin B$ with $B=\pm\tfrac{\pi}{4}$, where $\cos\tfrac{\pi}{4}=\sin\tfrac{\pi}{4}=\tfrac{\sqrt2}{2}$ and $\left(\sqrt2\right)\left(\tfrac{\sqrt2}{2}\right)=1$. The whole family of addition and double angle identities is set out with the trigonometric integrals in the next section; this member is the one this week needs, and the reason to keep it here is that a limit question hands you the sum, not the identity.
Check yourself
Close the page and write from memory: the derivative of an inverse at a point; the derivative of $a^{x}$; $\arcsin$ and $\arctan$ with their intervals; the hyperbolic identity with its sign; the two shapes the rule accepts; the three rewrites. Then check against the formula card and mark what you invented rather than recalled.
Given $f(2)=9$ and $f'(2)=3$, produce $\left(f^{-1}\right)'(9)$ and say why the derivative is evaluated at $2$?
c-inverse-slope
Differentiate $4^{x}$, $\log_3 x$ and $x^{\sqrt2}$ without confusing the three rules?
c-general-base
Derive $(\arcsin x)'$ from $\sin y=x$, including the sentence that fixes the sign of the root?
c-arcsin-arctan
Explain the minus sign in $(\arccos x)'$ without quoting a table, and state where each formula stops being valid?
c-arcsec-sign-family
Write $\cosh$ and $\sinh$ from memory, prove $\cosh^{2}-\sinh^{2}=1$ in one line, and recover $\sinh^{-1}x$ as a logarithm?
c-hyperbolic
Look at a quotient and say in one line whether the rule is allowed, including the case where it is allowed but useless?
c-lhopital
Take $\lim_{x\to 0^{+}}x^{x}$ from its shape to its value without help, and remember the last line?
c-recast-forms
Glossary (10 terms)
principal branchesas dal
The piece of a periodic function's domain on which it is one to one, chosen once so an inverse exists; for the sine, $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.
A function returning the angle on the principal branch whose sine, cosine or tangent is the given number; written $\arcsin$, $\arccos$, $\arctan$.
genel üstel fonksiyon
The function $a^{x}=e^{x\ln a}$ for a fixed base $a>0$; its derivative is itself multiplied by the constant $\ln a$.
general logarithmgenel logaritma
The function $\log_a x=\dfrac{\ln x}{\ln a}$, the inverse of $a^{x}$; its derivative is $\dfrac{1}{x\ln a}$.
ters fonksiyon teoremi
The statement that a one-to-one differentiable function with non-zero derivative has a differentiable inverse whose slope at an output is the reciprocal of the slope at the matching input.
hyperbolic functionhiperbolik fonksiyon
A combination of $e^{x}$ and $e^{-x}$ written $\sinh$, $\cosh$ or $\tanh$, satisfying $\cosh^{2}-\sinh^{2}=1$ and parametrising a hyperbola.
The inverse of a hyperbolic function; unusually it has an elementary closed form, since solving for it means solving a quadratic in $e^{y}$.
indeterminate formbelirsizlik
The symbol a substitution produces when it decides nothing, such as zero over zero; it reports that the expression must be rewritten, never what the answer is.
l'Hôpital's ruleL'Hôpital kuralı
The theorem that a limit of shape $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ equals the limit of the quotient of the derivatives, provided that second limit exists.
The standard move for a power form: take logarithms, evaluate the resulting product or quotient, and exponentiate at the end.
What comes next
§13 · Integration techniques: parts and trigonometric integrals
Every derivative built here gets read backwards next: the arcsine and arctangent formulas become the antiderivatives behind trigonometric substitution, and the product rule reversed becomes integration by parts.
Sources
James Stewart, Calculus, Ninth Edition Sections 6.4*, 6.6 and 6.8, which are the three named on this week's syllabus line.
Standard results assumed from earlier in this course The chain rule, implicit differentiation, the consequence that a zero derivative on an interval means a constant, and the derivatives of $\ln x$ and $e^{x}$.