03Derivatives: definition and basic differentiation rules

A claim of the form every falls to one counterexample. Ask which continuous graph you have already seen that misbehaves at a point.

Differentiability at $0$ needs $\lim_{h\to0}\frac{f(h)-f(0)}{h}$ to exist as a finite number; continuity only needs $\lim_{x\to0}f(x)=f(0)$.

Take $f(x)=\vert x\vert$: it is continuous at $0$, and its quotient there is $\frac{\vert h\vert}{h}$, which is $+1$ for $h>0$ and $-1$ for $h<0$.

So the claim is false, and $\vert x\vert$ at $0$ is the counterexample.

This function can be differentiated two ways. Find the route that needs no new rule at all, and use it as the referee.

$x^{2}\cdot x^{3}=x^{5}$, and the power rule gives $\frac{d}{dx}x^{5}=5x^{4}$. The product rule must agree: $(fg)'=f'g+fg'$.

Product rule with $f=x^{2}$, $g=x^{3}$: $2x\cdot x^{3}+x^{2}\cdot3x^{2}=2x^{4}+3x^{4}$.

Both routes give $5x^{4}$.

$f''(a)$ is a statement about the function $f'$, not about $f$. So ask what $f'$ has to be like near $a$, not what $f$ is like.

$f''(a)$ exists exactly when $f'$ is differentiable at $a$, and a function is differentiable only where it has no corner. So the question is whether $f'$ can have a corner.

Take $f(x)=x\vert x\vert$, which is $x^{2}$ for $x\ge0$ and $-x^{2}$ for $x<0$. Then $f'(x)=2\vert x\vert$ — a corner at $0$.

So the claim is false: $f'(0)=0$ exists, but $f''(0)$ does not.

The claim runs from $F$ to $f$, so nothing is being assumed about $f$ itself. The real question is how badly behaved the derivative of a differentiable function is allowed to be.

Differentiable is stronger than continuous. So one function that is a derivative everywhere and yet fails to be continuous at a single point settles the claim, and this section already carries such a function.

Take $F(x)=x^{2}\sin(1/x)$ with $F(0)=0$. The squeeze gives $F'(0)=0$, and away from the origin $F'(x)=2x\sin(1/x)-\cos(1/x)$.

That $F'$ has no limit at $0$ because of the $\cos(1/x)$ term, so it is not continuous there and therefore not differentiable there.

Each part is the same three moves: substitute, make $h$ a factor of the numerator, cancel. Decide in advance which move will do it for each function.

(a) expand $(3+h)^{2}$; (b) multiply by the conjugate $\sqrt{4+h}+2$; (c) put $\frac{1}{x+h+1}-\frac{1}{x+1}$ over one denominator. Then $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ in each case.

(a) $\frac{(3+h)^{2}-4(3+h)+1-(-2)}{h}=\frac{2h+h^{2}}{h}=2+h$. (b) $\frac{\sqrt{4+h}-2}{h}=\frac{1}{\sqrt{4+h}+2}$. (c) $\frac{-h}{h(x+h+1)(x+1)}=\frac{-1}{(x+h+1)(x+1)}$.

So $f_{1}'(3)=2$, $f_{2}'(1)=\frac14$, and $f_{3}'(x)=-\frac{1}{(x+1)^{2}}$.

Look at each expression and ask whether the power rule can even see it. If it cannot, the first line of your answer is a rewrite.

$\frac{d}{dx}x^{n}=nx^{n-1}$ with $\sqrt{x}=x^{1/2}$ and $\frac{2}{x^{3}}=2x^{-3}$; and $(fg)'=f'g+fg'$ for part (c).

(a) $p'=24x^{3}-\tfrac32x^{2}+7$, then differentiate again. (b) $q=3x^{1/2}-2x^{-3}$, so $q'=\tfrac32x^{-1/2}+6x^{-4}$. (c) expanded: $r=4x^{2}-4x+1$.

(a) $p''=72x^{2}-3x$; (b) $q'=\frac{3}{2\sqrt{x}}+\frac{6}{x^{4}}$; (c) $r'=8x-4$ by both routes.

Check first whether the fraction can be split — if it cannot, the quotient rule is unavoidable. Then remember what horizontal tangent says about $v'$.

$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$ with $f=2x+1$, $f'=2$, $g=x^{2}+x+1$, $g'=2x+1$. Horizontal tangent means $v'(x)=0$, and a fraction is zero exactly when its numerator is.

$$v'=\frac{2\left(x^{2}+x+1\right)-(2x+1)(2x+1)}{\left(x^{2}+x+1\right)^{2}}=\frac{2x^{2}+2x+2-4x^{2}-4x-1}{\left(x^{2}+x+1\right)^{2}}$$. Setting the numerator to zero: $-2x^{2}-2x+1=0$.

$v'(x)=\frac{1-2x-2x^{2}}{\left(x^{2}+x+1\right)^{2}}$, which is zero at $x=\frac{-1\pm\sqrt3}{2}$.

For (a) and (b) do not try to see the pattern before computing; compute three or four, then look.

Each differentiation is the power rule again. For (b) write $g(x)=x^{-1}$ so that every derivative is a single power.

(a) $f'=5x^{4}+12x^{2}$, $f''=20x^{3}+24x$, $f'''=60x^{2}+24$, $f^{(4)}=120x$, $f^{(5)}=120$. (b) $-x^{-2},\ 2x^{-3},\ -6x^{-4},\ 24x^{-5}$. (c) $k'=4x^{3}-6x^{2}$.

(a) $f^{(6)}=0$, because each derivative drops the degree by one and the degree was $5$. (b) $g^{(n)}(x)=\frac{(-1)^{n}n!}{x^{n+1}}$. (c) $k''(x)=12x^{2}-12x$, so $k''(1)=0$.

All three parts ask about slope, so the first line of the answer is the same in each: differentiate once and keep the result in front of you.

$y'=3x^{2}-6x$. A tangent line needs $y-f(a)=f'(a)(x-a)$; horizontal means $y'=0$; slope $9$ means $y'=9$.

(a) $f(1)=0$ and $f'(1)=3-6=-3$. (b) $3x^{2}-6x=3x(x-2)=0$. (c) $3x^{2}-6x=9\Rightarrow x^{2}-2x-3=0\Rightarrow(x-3)(x+1)=0$.

(a) $y=-3x+3$; (b) $(0,2)$ and $(2,-2)$; (c) $(3,2)$ and $(-1,-2)$.

Two unknowns need two equations. Ask what would go wrong at $x=1$ if only the slopes matched, and what would go wrong if only the values matched.

Differentiability at $1$ requires continuity at $1$ (Theorem 3.3) and equal one sided difference quotients. That is the pair of equations.

Continuity: $\lim_{x\to1^{-}}x^{2}=1$ must equal $\lim_{x\to1^{+}}(ax+b)=a+b$, so $a+b=1$. Slopes: the left branch has slope $2x\big\vert_{x=1}=2$, the right branch has slope $a$, so $a=2$.

From $a=2$ and $a+b=1$: $b=-1$.

Write down the three rules you will use before you look at any numbers. Then check that each rule's hypothesis is met by the table.

$(fg)'=f'g+fg'$; $\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$ (legal here since $g(2)=4\neq0$); $(cf\pm dg)'=cf'\pm dg'$.

With $f(2)=3,\ f'(2)=-1,\ g(2)=4,\ g'(2)=5$: (a) $(-1)(4)+(3)(5)$; (b) $\frac{(-1)(4)-(3)(5)}{4^{2}}$; (c) $5(-1)-2(5)$.

(a) $-4+15=11$; (b) $\frac{-19}{16}$; (c) $-5-10=-15$.

Compare the expression with $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ and ask what $f$ and $a$ would have to be for the two to match.

Matching term by term: $f(a+h)=\sqrt[3]{8+h}$ and $f(a)=2$ force $f(x)=x^{1/3}$ and $a=8$, since $\sqrt[3]{8}=2$. Then use $\frac{d}{dx}x^{n}=nx^{n-1}$.

$f'(x)=\frac13x^{-2/3}$, and at $x=8$: $8^{2/3}=\left(\sqrt[3]{8}\right)^{2}=4$.

So the limit is $f'(8)=\frac{1}{3\cdot4}=\frac{1}{12}$.

Differentiate first and look hard at the result before answering (b) and (c). The shape of $y'$ is the entire question.

$\frac{d}{dx}x^{n}=nx^{n-1}$ term by term; then recall that a squared real quantity is never negative, and that slope $3$ means $y'=3$.

$y'=3x^{2}+6x+3=3\left(x^{2}+2x+1\right)=3(x+1)^{2}$, which is $3$ times a square. For (c): $3(x+1)^{2}=3$.

(b) $y'\ge0$ always, with $y'=0$ only at $x=-1$; (c) $(x+1)^{2}=1$ gives $x=0$ and $x=-2$, so the points are $(0,3)$ and $(-2,1)$.

Substitution gives $\frac{0}{0}$, so something has to be rewritten — but before grinding, look at the shape of the expression and ask what it reminds you of.

Either factor $x^{5}-32$ as $(x-2)\left(x^{4}+2x^{3}+4x^{2}+8x+16\right)$, or notice the pattern $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$ with $f(x)=x^{5}$ and $a=2$, and use $\frac{d}{dx}x^{n}=nx^{n-1}$.

By the second route: $f'(x)=5x^{4}$, so the limit is $f'(2)=5\cdot2^{4}$. By the first, substituting $x=2$ into the long factor gives $16+16+16+16+16$.

Either way the limit is $80$.

Simplify the formula for $x\neq3$ before doing anything else; the whole problem changes appearance once you do.

Continuity at $3$ needs $\lim_{x\to3}f(x)=f(3)=k$. Then differentiability is decided by $\lim_{x\to3}\frac{f(x)-f(3)}{x-3}$.

$\frac{x^{2}-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq3$, so $\lim_{x\to3}f(x)=6$; with $k=6$ the rule $f(x)=x+3$ holds at $x=3$ as well.

So $k=6$, and $f$ is the line $y=x+3$, whose derivative is $1$ everywhere: $f'(3)=1$.

Write the quotient at $a=0$ using $f(0)=0$, and see what is left. Then ask which theorem from the limits chapter deals with a factor that refuses to settle.

The Squeeze Theorem: if $g\le F\le h$ near $0$ and $g,h$ both tend to $L$, then $F\to L$. Combine it with the bound $\left\vert\sin\left(\frac1h\right)\right\vert\le1$.

$\frac{f(h)-f(0)}{h}=\frac{h^{2}\sin\left(\frac1h\right)}{h}=h\sin\left(\frac1h\right)$, and $-\vert h\vert\le h\sin\left(\frac1h\right)\le\vert h\vert$ because the sine factor never leaves $[-1,1]$.

Both bounds tend to $0$, so the middle does too, and $f'(0)=0$.

Three numbers decide this question: $f'(0)$, $\lim_{x\to0^{-}}f'(x)$ and $\lim_{x\to0^{+}}f'(x)$. Before computing anything, ask which of the three a branch formula is entitled to tell you, and which one it is silent about.

At a point where the formula changes, $f'(0)$ exists exactly when the two one sided limits of $\frac{f(h)-f(0)}{h}$ agree. Continuity of $f'$ at $0$ is a different statement: it asks whether $\lim_{x\to0}f'(x)$ equals that number. Part (b) is the same definition run once more, with $f'$ in the role of $f$.

Right hand quotient: $\frac{h^{4/3}+h}{h}=h^{1/3}+1$. Left hand quotient: $\frac{h/(1-h)}{h}=\frac{1}{1-h}$. Away from $0$ the power rule gives $f'(x)=\frac{4}{3}x^{1/3}+1$ for $x>0$, and the quotient rule gives $f'(x)=\frac{1}{(1-x)^{2}}$ for $x<0$.

Both one sided quotients tend to $1$, and both branch formulas also tend to $1$, which settles (a). For (b) form $\frac{f'(h)-f'(0)}{h}$ on each side: on the right it is $\frac{4}{3}h^{-2/3}$.

Darboux's theorem constrains which values a derivative is obliged to hit. So the first job is not calculus at all: describe the set of values $\varphi$ actually takes on a short closed interval containing $0$.

Split $\lvert x\rvert$ by sign to remove it. A function fails the intermediate value property on $[a,b]$ as soon as you can name one number strictly between the two endpoint values that the function never equals anywhere on $[a,b]$.

For $x>0$ the quotient is $x-1$ and for $x<0$ it is $1-x$. On $[-1,1]$ this gives the values $1-x\in(1,2]$ for $x\in[-1,0)$, the single value $0$ at $x=0$, and $x-1\in(-1,0]$ for $x\in(0,1]$, with endpoints $\varphi(-1)=2$ and $\varphi(1)=0$.

Nothing on $[-1,1]$ is sent to $\tfrac12$, although $\tfrac12$ lies between the two endpoint values. For (b), redefining $\varphi(0)$ adds at most one number to that list, while a whole interval of values is missing.

Only one quantity is ever known here, and it is the limit of one very particular quotient. So (a) asks what has to happen to the expression on the page before that quotient becomes visible in it, while (b) and (c) ask what such an expression is capable of missing.

For fixed $k\neq0$, $\frac{f(2+kh)-f(2)}{h}=k\cdot\frac{f(2+kh)-f(2)}{kh}\to k\,f'(2)$, and adding and subtracting $f(2)$ splits a two sided increment into two such pieces. Existence of $f'(2)$ needs both one sided limits of $\frac{f(2+h)-f(2)}{h}$, so a quotient that samples one side only, or that lets the two sides cancel, can survive when $f'(2)$ does not.

In (a) write $\frac{f(2+3h)-f(2-h)}{h}=3\cdot\frac{f(2+3h)-f(2)}{3h}+\frac{f(2-h)-f(2)}{-h}$. For (b) test the cusp $f(x)=(x-2)^{2/3}$, whose two increments have the same value. For (c) note that $h^{2}>0$ for every $h\neq0$.

Part (a) is $3f'(2)+f'(2)$. In (b) both terms of the numerator equal $\lvert h\rvert^{2/3}$, so the quotient is identically zero. In (c) any function that is flat on the right of $2$ and has a different slope on the left defeats the claim.

Each part asks how much of $f$ survives an operation performed on it. Before hunting for examples, ask what a function with only two values can do to the expression $t^{2}+t$, what squaring does to a sign, and what the hypothesis in (c) is entitled to say about the number $f(0)$ itself.

A function taking only the two values $\alpha\neq\beta$, each on a dense set, is discontinuous at every point, because then every interval meets both sets on which it is constant. If a polynomial $p$ satisfies $p(\alpha)=p(\beta)$, then $p(f(x))$ is constant. And differentiability at $0$ is a statement about $\frac{f(x)-f(0)}{x}$, so it needs the value $f(0)$, which a limit as $x\to0$ never inspects.

For (a) solve $t^{2}+t=0$ and read off the two roots. For (b) try $f(x)=\lvert x\rvert$, whose square is $x^{2}$. For (c) try $f(x)=x\lvert x\rvert$ for $x\neq0$ together with $f(0)=1$.

In (a) let $f$ be $-1$ on $\mathbb{Q}$ and $0$ elsewhere. In (b) the square is smooth while the original has a corner. In (c) the claim dies on the value at $0$, and continuity revives it because it forces $f(0)=\lim_{x\to0}x\cdot\frac{f(x)}{x}=0$.

Nothing here is a differentiate this question. In (a) you get to choose the $x$; in (b) the identity holds at every $x$ at once, so you may choose several; in (c) nothing is chosen at all — everything the equation names has to be computed and put in.

An identity true for every $x$ stays true at any single $x$ you substitute. Two unknown values need two equations, so in (a) look for the input where $\sqrt{x}$ and $x$ are the same number. In (b) use that $\sin 3x$ vanishes at $x=0$ while $\cos 3x$ vanishes at $x=\pi/6$.

(a) At $x=1$ the identity reads $f(1)+2f(1)=4$. (b) $x=0$ leaves $3A-B-1=0$ and $x=\pi/6$ leaves $A+2B-5=0$. (c) Rewrite $y=x+3x^{-1}$, so that $y'=1-3x^{-2}$ and $y''=6x^{-3}$.

(a) $3f(1)=4$. (b) The two linear equations give one pair of constants. (c) The substituted terms cancel in two groups and leave $0$.