Exercise 0.1: The derivative at a point from the difference quotient

For each function, compute $f'(a)$ at the given point using ONLY the definition $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ . No differentiation rule is allowed: write the difference quotient, simplify it until the $h$ in the denominator disappears, then take the limit.

Three moves are enough, one per family: expand and then factor out $h$ for a polynomial, multiply by the conjugate for a square root, use a common denominator for a fraction.

a) $f(x)=3x+5$ at $a=2$ .
b) $f(x)=x^{2}-4x$ at $a=3$ .
c) $f(x)=2x^{2}+x-1$ at $a=-1$ .
d) $f(x)=x^{3}$ at $a=2$ .
e) $f(x)=\sqrt{x+5}$ at $a=4$ .
f) $f(x)=\sqrt{2x+1}$ at $a=4$ .
g) $f(x)=\frac{3}{x}$ at $a=-1$ .
h) $f(x)=\frac{1}{x-2}$ at $a=3$ .
i) $f(x)=\frac{x}{x+1}$ at $a=1$ .
j) $f(x)=\frac{1}{\sqrt{x}}$ at $a=4$ .

Type your answers, the page tells you right or wrong 0/10

a)

f'(2)

b)

f'(3)

c)

f'(-1)

d)

f'(2)

e)

f'(4)

f)

f'(4)

g)

f'(-1)

h)

f'(3)

i)

f'(1)

j)

f'(4)

Show the solution

Answers

a) $f'(2)=3$
b) $f'(3)=2$
c) $f'(-1)=-3$
d) $f'(2)=12$
e) $f'(4)=\frac{1}{6}$
f) $f'(4)=\frac{1}{3}$
g) $f'(-1)=-3$
h) $f'(3)=-1$
i) $f'(1)=\frac{1}{4}$
j) $f'(4)=-\frac{1}{16}$

The plan is the same on every line: compute $f(a)$ , then $f(a+h)$ , write the difference, divide it by $h$ after simplifying it, and finally let $h$ tend to 0.

As long as the $h$ in the denominator is there, replacing $h$ by 0 gives $\frac{0}{0}$ : the whole simplification exists to make it cancel. You are allowed to divide by $h$ because, inside a limit, $h\ne 0$ .

a)

$f(2)=11$ and $f(2+h)=3(2+h)+5=11+3h$ .

\begin{aligned}\frac{f(2+h)-f(2)}{h} &= \frac{11+3h-11}{h} \\ &= \frac{3h}{h} = 3\end{aligned}

The quotient is constant, so $f'(2)=3$ .

For a linear function, the difference quotient does not even depend on $h$ : the secant line is the line, and the derivative is its slope, everywhere.

b)

$f(3)=9-12=-3$ and:

\begin{aligned}f(3+h) &= (3+h)^{2}-4(3+h) \\ &= 9+6h+h^{2}-12-4h \\ &= -3+2h+h^{2}\end{aligned}

\begin{aligned}\frac{f(3+h)-f(3)}{h} &= \frac{2h+h^{2}}{h} \\ &= \frac{h(2+h)}{h} \\ &= 2+h \to 2\end{aligned}

So $f'(3)=2$ .

Pitfall: forgetting the $-4h$ when expanding $-4(3+h)$ ; the quotient becomes $6+h$ and the “derivative” 6 instead of 2.

c)

$f(-1)=2-1-1=0$ and:

\begin{aligned}&f(-1+h) \\ &\quad = 2(1-2h+h^{2})+(-1+h)-1 \\ &\quad = -3h+2h^{2}\end{aligned}

\begin{aligned}\frac{f(-1+h)-f(-1)}{h} &= \frac{-3h+2h^{2}}{h} \\ &= -3+2h \to -3\end{aligned}

So $f'(-1)=-3$ .

Pitfall: the sign in $(-1+h)^{2}=1-2h+h^{2}$ . Writing it $1+2h+h^{2}$ gives $f'(-1)=5$ , wrong and even of the wrong sign.

d)

Expand the cube:

\begin{aligned}(2+h)^{3} &= 8+12h+6h^{2}+h^{3} \\ f(2+h)-f(2) &= 12h+6h^{2}+h^{3}\end{aligned}

\displaystyle \frac{f(2+h)-f(2)}{h}=12+6h+h^{2}\to 12

So $f'(2)=12$ .

Pitfall: $(2+h)^{3}\ne 8+h^{3}$ . With that error the quotient is $h^{2}$ and the derivative 0, which a curve this steep at 2 contradicts at a glance.

e)

$f(4)=\sqrt{9}=3$ . Multiply top and bottom by the conjugate $\sqrt{9+h}+3$ :

\begin{aligned}\frac{\sqrt{9+h}-3}{h} &= \frac{(9+h)-9}{h(\sqrt{9+h}+3)} \\ &= \frac{h}{h(\sqrt{9+h}+3)} \\ &= \frac{1}{\sqrt{9+h}+3} \to \frac{1}{6}\end{aligned}

So $f'(4)=\frac{1}{6}$ .

The conjugate is not there to “get rid of the root”: it turns a difference of roots into a difference of squares, in which $h$ appears as a factor.

f)

$f(4)=\sqrt{9}=3$ and $f(4+h)=\sqrt{9+2h}$ . Conjugate $\sqrt{9+2h}+3$ :

\begin{aligned}&\frac{\sqrt{9+2h}-3}{h} \\ &\quad = \frac{(9+2h)-9}{h(\sqrt{9+2h}+3)} \\ &\quad = \frac{2}{\sqrt{9+2h}+3} \to \frac{1}{3}\end{aligned}

So $f'(4)=\frac{1}{3}$ .

Pitfall: writing $f(4+h)=\sqrt{2\cdot 4+1+h}$ . It is $x$ that is replaced by $4+h$ , so $h$ is multiplied by 2 just like $x$ . The error gives $\frac{1}{6}$ , half the correct value.

g)

$f(-1)=-3$ and $f(-1+h)=\frac{3}{h-1}$ .

\begin{aligned}&f(-1+h)-f(-1) \\ &\quad = \frac{3}{h-1}+3 \\ &\quad = \frac{3+3(h-1)}{h-1} = \frac{3h}{h-1}\end{aligned}

\displaystyle \frac{f(-1+h)-f(-1)}{h}=\frac{3}{h-1}\to\frac{3}{-1}=-3

So $f'(-1)=-3$ . Sign check: $\frac{3}{x}$ decreases on each branch, so a negative derivative is expected.

Pitfall: subtracting $3$ instead of $-3$ . The numerator is then $\frac{6-3h}{h-1}$ , which tends to $-6$ and not to 0: that is the alarm bell, because in a derivative the difference $f(a+h)-f(a)$ always tends to 0.

h)

$f(3)=1$ and $f(3+h)=\frac{1}{1+h}$ . Over a common denominator:

\begin{aligned}\frac{1}{1+h}-1 &= \frac{1-(1+h)}{1+h} \\ &= \frac{-h}{1+h}\end{aligned}

\displaystyle \frac{f(3+h)-f(3)}{h}=\frac{-1}{1+h}\to -1

So $f'(3)=-1$ .

Pitfall: dropping the parentheses, $1-1+h=h$ , which gives $+1$ .

i)

A mix of polynomial and fraction. $f(1)=\frac{1}{2}$ and $f(1+h)=\frac{1+h}{2+h}$ .

\begin{aligned}\frac{1+h}{2+h}-\frac{1}{2} &= \frac{2(1+h)-(2+h)}{2(2+h)} \\ &= \frac{h}{2(2+h)}\end{aligned}

\displaystyle \frac{f(1+h)-f(1)}{h}=\frac{1}{2(2+h)}\to\frac{1}{4}

So $f'(1)=\frac{1}{4}$ .

Pitfall: believing that the derivative of a quotient is the quotient of the derivatives, $\frac{1}{1}=1$ ; the definition leaves no room for that shortcut.

j)

A mix of fraction and square root, the two moves one after the other. $f(4)=\frac{1}{2}$ .

Common denominator first:

\displaystyle \frac{1}{\sqrt{4+h}}-\frac{1}{2}=\frac{2-\sqrt{4+h}}{2\sqrt{4+h}}

Then the conjugate, $2+\sqrt{4+h}$ : the numerator becomes $4-(4+h)=-h$ , which gives the quotient:

\displaystyle \frac{-1}{2\sqrt{4+h}\,(2+\sqrt{4+h})}\to\frac{-1}{2\cdot 2\cdot 4}=-\frac{1}{16}

So $f'(4)=-\frac{1}{16}$ .

The order matters: combine the fractions before looking for the conjugate, otherwise there is no difference of roots to multiply.

Overall check, with no rule but with a calculator: for $h=0.001$ , the difference quotient on each line falls within one hundredth of the value found. It is the best check of a derivative from the definition, and it takes ten seconds.

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Exercise 0.2: The derivative function from the difference quotient

For each function, compute $f'(x)$ for every $x$ where it exists, using ONLY the definition $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ . Same method as in exercise 0.1, but $x$ stays a letter: only $h$ tends to 0, and the result is a FUNCTION.

Once the computation is done, choose the formula you obtained from the list. For the square roots and the fractions, also state for which $x$ the derivative exists.

a) $f(x)=5x-7$ .
b) $f(x)=x^{2}+3x$ .
c) $f(x)=3x^{2}-2x+1$ .
d) $f(x)=x^{3}-x$ .
e) $f(x)=\sqrt{x-3}$ .
f) $f(x)=\sqrt{5-x}$ .
g) $f(x)=\frac{2}{x}$ .
h) $f(x)=\frac{1}{x+3}$ .
i) $f(x)=\frac{x+1}{x-1}$ .
j) $f(x)=\frac{1}{x^{2}}$ .

Type your answers, the page tells you right or wrong 0/11

a)

f'(x)

b)

f'(x)

c)

f'(x)

d)

f'(x)

e)

f'(x)

Differentiable for

f)

f'(x)

g)

f'(x)

h)

f'(x)

i)

f'(x)

j)

f'(x)

Show the solution

Answers

a) $f'(x)=5$ for every real $x$
b) $f'(x)=2x+3$
c) $f'(x)=6x-2$
d) $f'(x)=3x^{2}-1$
e) $f'(x)=\frac{1}{2\sqrt{x-3}}$ for $x>3$
f) $f'(x)=-\frac{1}{2\sqrt{5-x}}$ for $x<5$
g) $f'(x)=-\frac{2}{x^{2}}$ for $x\ne 0$
h) $f'(x)=-\frac{1}{(x+3)^{2}}$ for $x\ne -3$
i) $f'(x)=-\frac{2}{(x-1)^{2}}$ for $x\ne 1$
j) $f'(x)=-\frac{2}{x^{3}}$ for $x\ne 0$

The difference from exercise 0.1 fits in one sentence: $x$ is fixed during the whole computation, only $h$ moves.

So expand $f(x+h)$ keeping $x$ as a letter, and replace $h$ by 0 only on the very last line. Each result can be checked by plugging in a value of $x$ and comparing with the quotient computed on a calculator for a small $h$ .

a)

\begin{aligned}&f(x+h)-f(x) \\ &\quad = 5(x+h)-7-(5x-7) \\ &\quad = 5h\end{aligned}

The quotient equals $5$ , so $f'(x)=5$ for every real $x$ .

The constant $-7$ disappears in the difference: it shifts the line without changing its slope.

b)

\displaystyle f(x+h)=x^{2}+2xh+h^{2}+3x+3h

\begin{aligned}&f(x+h)-f(x) \\ &\quad = 2xh+h^{2}+3h \\ &\quad = h(2x+h+3)\end{aligned}

Quotient: $2x+h+3\to 2x+3$ . So $f'(x)=2x+3$ .

Pitfall: stopping at $2x+3+h$ . As long as $h$ is still there, you have only computed the slope of a secant line, not the derivative.

c)

$3(x+h)^{2}=3x^{2}+6xh+3h^{2}$ and $-2(x+h)=-2x-2h$ .

\begin{aligned}f(x+h)-f(x) &= 6xh+3h^{2}-2h \\ &= h(6x+3h-2)\end{aligned}

Quotient: $6x+3h-2\to 6x-2$ . So $f'(x)=6x-2$ .

Check: at $x=0$ , $f'(0)=-2$ , and the quotient $\frac{f(0.001)-f(0)}{0.001}=-1.997$ .

Pitfall: keeping the constant $+1$ ; it cancels in $f(x+h)-f(x)$ as in a).

d)

\displaystyle (x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}

\begin{aligned}&f(x+h)-f(x) \\ &\quad = 3x^{2}h+3xh^{2}+h^{3}-h \\ &\quad = h(3x^{2}+3xh+h^{2}-1)\end{aligned}

Quotient: $3x^{2}+3xh+h^{2}-1\to 3x^{2}-1$ . So $f'(x)=3x^{2}-1$ .

The terms in $h$ and $h^{2}$ of the quotient are exactly the ones that vanish in the limit: an incomplete expansion of $(x+h)^{3}$ therefore does not always show in the result, which is why you should write it out in full.

e)

Multiply by the conjugate $\sqrt{x+h-3}+\sqrt{x-3}$ : the numerator becomes $(x+h-3)-(x-3)=h$ .

\begin{aligned}&\frac{\sqrt{x+h-3}-\sqrt{x-3}}{h} \\ &\quad = \frac{1}{\sqrt{x+h-3}+\sqrt{x-3}} \\ &\quad \to \frac{1}{2\sqrt{x-3}}\end{aligned}

So $f'(x)=\frac{1}{2\sqrt{x-3}}$ for $x>3$ .

At $x=3$ , $f$ is defined but the derivative is not: the denominator vanishes, and the tangent line there is vertical.

Pitfall: forgetting the 2, which comes from the fact that both roots in the denominator tend to $\sqrt{x-3}$ .

f)

The conjugate $\sqrt{5-x-h}+\sqrt{5-x}$ gives the numerator $(5-x-h)-(5-x)=-h$ .

\begin{aligned}&\frac{\sqrt{5-x-h}-\sqrt{5-x}}{h} \\ &\quad = \frac{-1}{\sqrt{5-x-h}+\sqrt{5-x}} \\ &\quad \to \frac{-1}{2\sqrt{5-x}}\end{aligned}

So $f'(x)=-\frac{1}{2\sqrt{5-x}}$ for $x<5$ .

Graphical check: $\sqrt{5-x}$ decreases as $x$ increases, so a positive derivative would be absurd.

Pitfall: losing the minus sign, which comes from $-(x+h)=-x-h$ .

g)

Common denominator:

\begin{aligned}\frac{2}{x+h}-\frac{2}{x} &= \frac{2x-2(x+h)}{x(x+h)} \\ &= \frac{-2h}{x(x+h)}\end{aligned}

Quotient: $\frac{-2}{x(x+h)}\to-\frac{2}{x^{2}}$ . So $f'(x)=-\frac{2}{x^{2}}$ for $x\ne 0$ .

Check: at $x=-1$ , you get $-2$ , consistent with part g) of exercise 0.1, which gave $-3$ for $\frac{3}{x}$ at the same point.

h)

\begin{aligned}&\frac{1}{x+h+3}-\frac{1}{x+3} \\ &\quad = \frac{(x+3)-(x+h+3)}{(x+3)(x+h+3)} \\ &\quad = \frac{-h}{(x+3)(x+h+3)}\end{aligned}

Quotient: $\frac{-1}{(x+3)(x+h+3)}\to-\frac{1}{(x+3)^{2}}$ . So $f'(x)=-\frac{1}{(x+3)^{2}}$ for $x\ne -3$ .

The square in the denominator is not a memorized formula: it is the product of two factors that become equal as $h\to 0$ .

i)

A mix of polynomial and fraction. Over a common denominator, the numerator of $f(x+h)-f(x)$ is:

\begin{aligned}&(x+h+1)(x-1)-(x+1)(x+h-1) \\ &\quad = (x^{2}+hx-h-1) \\ &\quad \quad -(x^{2}+hx+h-1) \\ &\quad = -2h\end{aligned}

Everything else cancels, which is the sign of a correct computation.

Quotient: $\frac{-2}{(x+h-1)(x-1)}\to-\frac{2}{(x-1)^{2}}$ . So $f'(x)=-\frac{2}{(x-1)^{2}}$ for $x\ne 1$ .

Pitfall: dividing the derivative of the top by that of the bottom, $\frac{1}{1}=1$ , wrong as soon as you test a value.

j)

\begin{aligned}\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}} &= \frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}} \\ &= \frac{-2xh-h^{2}}{x^{2}(x+h)^{2}}\end{aligned}

Quotient: $\frac{-(2x+h)}{x^{2}(x+h)^{2}}\to\frac{-2x}{x^{4}}=-\frac{2}{x^{3}}$ . So $f'(x)=-\frac{2}{x^{3}}$ for $x\ne 0$ .

Notice the sign: $-\frac{2}{x^{3}}$ is positive for $x<0$ , and $\frac{1}{x^{2}}$ does increase as $x$ rises towards 0 from the left.

Pitfall: $(x+h)^{2}=x^{2}+2xh+h^{2}$ , not $x^{2}+h^{2}$ ; the error leaves $-h^{2}$ in the numerator and a “derivative” equal to zero.

Summary of the two calculation exercises: the ten derivative functions found here are given in one line by the power and quotient rules of exercise 2.

The definition proves the rules; the rules do not excuse you from knowing how to redo the definition, which 201-SN2-RE exams ask for on at least one question almost every time.

Exercise 1: From the secant line to the tangent line: the definition

The figure shows $y=x^{2}$ at the point $A(1,1)$ , with two secant lines (in grey, the steeper one for $h=1$ ): as the second point slides towards $A$ , the secant line tips over towards the tangent line, which part b) asks you to find.

a) Compute the slope of the secant line joining $A(1,1)$ to the point with $x$ -coordinate $1+h$ for $h=1$ and then $h=0.5$ , and check your values on the figure.
b) Write the rate of change for an arbitrary $h$ , simplify it, then take the limit to obtain $f'(1)$ . Give the equation of the tangent line.
c) Compute $g'(2)$ from the DEFINITION for $g(x)=\frac{1}{x}$ .
d) Compute $k'(4)$ from the DEFINITION for $k(x)=\sqrt{x}$ . Which technique from the chapter on limits do you need to bring back, and why is the derivative always an indeterminate form $\frac{0}{0}$ ?

Show the solution

Answers

a) $3$ and $2.5$
b) $f'(1)=2$ ; tangent line $y=2x-1$
c) $g'(2)=-\frac{1}{4}$
d) $k'(4)=\frac{1}{4}$ with the conjugate

a)

For $h=1$ , the second point is $(2,4)$ : slope $\frac{4-1}{2-1}=3$ .

For $h=0.5$ , the point is $(1.5, 2.25)$ : slope $\frac{2.25-1}{0.5}=2.5$ .

These are indeed the two grey lines in the figure, flatter and flatter: the secant line approaches the tangent line from above.

b)

Rate of change, which may be simplified because $h\ne 0$ inside a limit:

\begin{aligned}\frac{f(1+h)-f(1)}{h} &= \frac{(1+h)^{2}-1}{h} \\ &= \frac{1+2h+h^{2}-1}{h} \\ &= \frac{h(2+h)}{h} = 2+h\end{aligned}

\displaystyle f'(1)=\lim_{h\to 0}(2+h)=2

Consistency: $h=1$ gave 3, $h=0.5$ gave 2.5, the trend towards 2 is clear.

Tangent line, the blue line:

\displaystyle y=f(1)+f'(1)(x-1)=1+2(x-1)=2x-1

c)

Combine the fractions in the numerator, then divide by $h$ :

\begin{aligned}\frac{g(2+h)-g(2)}{h} &= \frac{\frac{1}{2+h}-\frac{1}{2}}{h} \\ &= \frac{2-(2+h)}{2(2+h)}\cdot\frac{1}{h} \\ &= \frac{-1}{2(2+h)} \to -\frac{1}{4}\end{aligned}

So $g'(2)=-\frac{1}{4}$ : negative, as it should be for a decreasing hyperbola.

d)

Multiply by the conjugate $\sqrt{4+h}+2$ :

\begin{aligned}\frac{\sqrt{4+h}-2}{h} &= \frac{(4+h)-4}{h(\sqrt{4+h}+2)} \\ &= \frac{h}{h(\sqrt{4+h}+2)} \\ &= \frac{1}{\sqrt{4+h}+2} \to \frac{1}{4}\end{aligned}

So $k'(4)=\frac{1}{4}$ .

Why always $\frac{0}{0}$ : in the numerator, $f(a+h)-f(a)\to 0$ as soon as $f$ is continuous at $a$ ; in the denominator, $h\to 0$ by construction.

The derivative is defined as an indeterminate form: giving it a meaning is precisely why the chapter on limits exists, and each of its techniques (factoring, conjugate, common denominator) shows up here in the service of the derivative.

Exercise 2: The rules: power, product, quotient

The rules spare you from going back to the definition every time; you still need to know which one applies, and in which order.

a) Differentiate $f(x)=3x^{5}-2x^{3}+7x-4$ , then $g(x)=4\sqrt{x}+\frac{3}{x^{2}}$ (rewrite with exponents first).
b) Differentiate $h(x)=(x^{2}+1)(x^{3}-2x)$ with the product rule, then check by expanding first. What do you conclude?
c) Differentiate $u(x)=\frac{2x+1}{x-3}$ with the quotient rule, and determine the sign of $u'$ on its domain. What does this say about the function?
d) Prove the product rule from the definition, using the trick of adding and subtracting $f(x+h)g(x)$ .

Show the solution

Answers

a) $15x^{4}-6x^{2}+7$ ; $\frac{2}{\sqrt{x}}-\frac{6}{x^{3}}$
b) $h'(x)=5x^{4}-3x^{2}-2$ by both routes
c) $u'(x)=\frac{-7}{(x-3)^{2}}<0$ : decreasing on each branch
d) $(fg)'=f'g+fg'$

a)

Power rule term by term, the constant $-4$ disappears:

\displaystyle f'(x)=15x^{4}-6x^{2}+7

For $g$ , first rewrite with exponents, $g(x)=4x^{1/2}+3x^{-2}$ :

\begin{aligned}g'(x) &= 4\cdot\frac{1}{2}x^{-1/2}+3(-2)x^{-3} \\ &= \frac{2}{\sqrt{x}}-\frac{6}{x^{3}}\end{aligned}

Reflex: roots and fractions are converted to exponents before any differentiation; the power rule holds for every real exponent.

b)

With the product rule:

\begin{aligned}&h'(x) \\ &\quad = (2x)(x^{3}-2x)+(x^{2}+1)(3x^{2}-2) \\ &\quad = 2x^{4}-4x^{2}+3x^{4}-2x^{2}+3x^{2}-2 \\ &\quad = 5x^{4}-3x^{2}-2\end{aligned}

Check by expanding first: $h(x)=x^{5}-x^{3}-2x$ , so $h'=5x^{4}-3x^{2}-2$ ✓.

Conclusion: both routes give the same result, but expanding first is only possible for polynomials; the product rule works on any product ( $xe^{x}$ , $x\sin x$ ).

Pitfall: believing that $(fg)'=f'g'$ . Here $f'g'=(2x)(3x^{2}-2)=6x^{3}-4x$ , which has nothing to do with it.

c)

With the quotient rule:

\begin{aligned}u' &= \frac{2(x-3)-(2x+1)(1)}{(x-3)^{2}} \\ &= \frac{2x-6-2x-1}{(x-3)^{2}} \\ &= \frac{-7}{(x-3)^{2}}\end{aligned}

The denominator is a square, strictly positive everywhere except at 3: $u'<0$ on the whole domain.

The function is therefore strictly decreasing on $(-\infty,3)$ and on $(3,+\infty)$ : decreasing on each branch, but not “on its domain” as a single piece, since the asymptote at 3 separates two pieces between which no comparison makes sense.

d)

From the definition:

\displaystyle (fg)'(x)=\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}

Add and subtract $f(x+h)g(x)$ in the numerator, then group the terms:

\begin{aligned}&f(x+h)g(x+h)-f(x+h)g(x) \\ &\quad \quad +f(x+h)g(x)-f(x)g(x) \\ &\quad = f(x+h)\left[g(x+h)-g(x)\right] \\ &\quad \quad +g(x)\left[f(x+h)-f(x)\right]\end{aligned}

Divide by $h$ and take the limit: the first term tends to $f(x)g'(x)$ (because $f(x+h)\to f(x)$ , differentiability implying continuity), the second to $g(x)f'(x)$ .

\displaystyle (fg)'=f'g+fg'

The add-and-subtract trick is the hidden framework of most of the proofs in the course.

Exercise 3: The chain rule: differentiating compositions

The chain rule differentiates composite functions: peel the function like an onion, from the outside in, multiplying at each layer.

a) State the chain rule in both notations, then differentiate $f(x)=(3x^{2}+1)^{5}$ .
b) Differentiate $g(x)=\sqrt{x^{2}+9}$ and $h(x)=\sin(x^{3})$ . Be careful not to confuse $\sin(x^{3})$ with $\sin^{3}(x)$ : differentiate the latter as well.
c) Differentiate $k(x)=\left[(x^{2}+1)^{3}+2\right]^{4}$ : three nested layers.
d) A spherical balloon is being inflated: its radius grows at $2$ cm/s. Using the chain rule, find how fast the volume is growing when $r=10$ cm. Which reading of the chain rule does this problem illustrate?

Show the solution

Answers

a) $30x(3x^{2}+1)^{4}$
b) $\frac{x}{\sqrt{x^{2}+9}}$ ; $3x^{2}\cos(x^{3})$ ; $3\sin^{2}x\cos x$
c) $24x(x^{2}+1)^{2}\left[(x^{2}+1)^{3}+2\right]^{3}$
d) $800\pi\approx 2513$ cm³/s

a)

Lagrange notation: $[f(g(x))]'=f'(g(x))\cdot g'(x)$ .

Leibniz notation: $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$ , where the $du$ seem to “cancel”, which makes an excellent memory aid.

For $f(x)=(3x^{2}+1)^{5}$ : the outside is the fifth power, the inside $3x^{2}+1$ , whose derivative is $6x$ .

\begin{aligned}f'(x) &= 5(3x^{2}+1)^{4}\cdot 6x \\ &= 30x(3x^{2}+1)^{4}\end{aligned}

b)

$g(x)=(x^{2}+9)^{1/2}$ :

\begin{aligned}g'(x) &= \frac{1}{2}(x^{2}+9)^{-1/2}\cdot 2x \\ &= \frac{x}{\sqrt{x^{2}+9}}\end{aligned}

$h(x)=\sin(x^{3})$ : the outside is the sine, the inside $x^{3}$ .

\displaystyle h'(x)=\cos(x^{3})\cdot 3x^{2}=3x^{2}\cos(x^{3})

On the other hand, $\sin^{3}(x)=[\sin x]^{3}$ has the cube as its outside and the sine as its inside:

\displaystyle \left(\sin^{3}x\right)'=3\sin^{2}(x)\cos x

Pitfall: two very similar notations, two functions and two unrelated derivatives. Read the order of composition before differentiating.

c)

Three layers: the fourth power (outside), then $(x^{2}+1)^{3}+2$ , then $x^{2}+1$ . Peel them:

\begin{aligned}&k'(x) \\ &\quad = 4\left[(x^{2}+1)^{3}+2\right]^{3}\cdot\left[(x^{2}+1)^{3}+2\right]' \\ &\quad = 4\left[(x^{2}+1)^{3}+2\right]^{3}\cdot 3(x^{2}+1)^{2}\cdot 2x \\ &\quad = 24x(x^{2}+1)^{2}\left[(x^{2}+1)^{3}+2\right]^{3}\end{aligned}

Each layer adds a factor, and the process stops when the inside is finally an elementary expression.

d)

$V=\frac{4}{3}\pi r^{3}$ , where $r$ depends on time:

\displaystyle \frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}=4\pi r^{2}\cdot\frac{dr}{dt}

With $r=10$ and $\frac{dr}{dt}=2$ :

$\frac{dV}{dt}=4\pi(100)(2)=800\pi\approx 2513$ cm³/s.

The chain rule is the law of composition of rates: the volume does not depend directly on time, but on the radius, which itself depends on time, and the two rates multiply.

Notice the factor $4\pi r^{2}$ , the surface area of the sphere: the volume grows in proportion to the surface (you add a film of thickness $dr$ over the whole surface), which leads straight into the chapter on related rates.

Exercise 4: Higher-order derivatives

Differentiating a derivative gives a new function, and the operation can be repeated. The second derivative measures the change in the rate of change, a notion without which neither concavity nor acceleration would exist.

a) For $f(x)=x^{4}-3x^{2}+2x$ , compute $f'$ , $f''$ , $f'''$ , $f^{(4)}$ and $f^{(5)}$ . What do you notice for a polynomial?
b) An object has position $s(t)=t^{3}-6t^{2}+9t$ (in metres, $t$ in seconds). Find its velocity and its acceleration, then the times when it stops and the time when its acceleration is zero.
c) For $f(x)=\frac{1}{x}$ , compute $f'$ , $f''$ , $f'''$ , then conjecture a formula for $f^{(n)}(x)$ . Check it for $n=1$ and $n=2$ .
d) Give $f^{(n)}$ for $f(x)=e^{2x}$ , then for $f(x)=\sin x$ . How do these two behaviours differ radically?

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Answers

a) $4x^{3}-6x+2$ , $12x^{2}-6$ , $24x$ , $24$ , $0$
b) $v=3(t-1)(t-3)$ , $a=6t-12$ ; stops at 1 s and 3 s; $a=0$ at 2 s
c) $f^{(n)}(x)=(-1)^{n}n!\,x^{-(n+1)}$
d) $2^{n}e^{2x}$ ; the sine has a cycle of 4

a)

Differentiate repeatedly:

\begin{aligned}f'(x) &= 4x^{3}-6x+2 \\ f''(x) &= 12x^{2}-6 \\ f'''(x) &= 24x \\ f^{(4)}(x) &= 24 \\ f^{(5)}(x) &= 0\end{aligned}

Each differentiation lowers the degree by one: the $(n+1)$ -th derivative of a polynomial of degree $n$ is identically zero, and so are all the following ones. Here the degree is 4, and it is indeed $f^{(5)}$ that vanishes.

That is what makes polynomials so convenient: their sequence of derivatives stops.

b)

\begin{aligned}v(t)=s'(t) &= 3t^{2}-12t+9 \\ &= 3(t-1)(t-3) \\ a(t)=v'(t) &= 6t-12\end{aligned}

The object stops when $v(t)=0$ , that is at $t=1$ s and $t=3$ s. The acceleration is zero at $t=2$ s.

Reading the motion: from $t=0$ to $t=1$ the velocity is positive (the object moves forward), between $t=1$ and $t=3$ it is negative (it moves backward, with $v(2)=-3$ m/s), then it becomes positive again.

The instant $t=2$ , where $a=0$ , is when the object moves backward the fastest: it is the minimum of the velocity, not a stop.

Pitfall: confusing “zero acceleration” with “zero velocity”, the classic error in this type of question.

c)

Rewriting $f(x)=x^{-1}$ :

\begin{aligned}f'(x) &= -x^{-2} \\ f''(x) &= 2x^{-3} \\ f'''(x) &= -6x^{-4}\end{aligned}

The signs alternate, and the coefficients $1,2,6$ are the factorials $1!,2!,3!$ . Conjecture:

\displaystyle f^{(n)}(x)=(-1)^{n}\,n!\,x^{-(n+1)}

Check: for $n=1$ , $(-1)^{1}(1!)x^{-2}=-x^{-2}$ ✓; for $n=2$ , $(-1)^{2}(2!)x^{-3}=2x^{-3}$ ✓.

Unlike the polynomial, the sequence never stops: each differentiation pushes the negative exponent further down instead of lowering a degree.

d)

For $f(x)=e^{2x}$ , each differentiation brings out a factor of 2 through the chain rule:

\displaystyle f'(x)=2e^{2x},\quad f''(x)=4e^{2x},\quad f^{(n)}(x)=2^{n}e^{2x}

The function reproduces itself up to a factor, and that factor grows geometrically.

For $f(x)=\sin x$ :

\displaystyle f'=\cos x,\quad f''=-\sin x,\quad f'''=-\cos x,\quad f^{(4)}=\sin x

The cycle starts over: $f^{(n)}$ takes only four values, depending on the remainder of $n$ divided by 4.

The difference is radical: the exponential amplifies without end, the sine goes round in circles without ever growing. It is this period of four that will produce, in the integral calculus course, the Maclaurin series of the sine with its alternating terms and odd exponents.

Exercise 5: Tangent and normal lines

The derivative is not just a number: it is the slope of a specific line. Being able to write the equation of that line, and of its perpendicular, is required on every exam.

a) For $f(x)=x^{2}-4x+3$ , write the equation of the tangent line and of the normal line at the point with $x=3$ .
b) At which point is the tangent line to this curve PARALLEL to the line $y=8x+5$ ? Give its equation.
c) Find all the tangent lines to the parabola $y=x^{2}$ that pass through the point $P(0,-1)$ , which lies outside the curve. How many are there?
d) Why is the slope of the normal line equal to $-\frac{1}{f'(a)}$ ? What happens to the normal line when $f'(a)=0$ ?

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Answers

a) Tangent $y=2x-6$ ; normal $y=-\frac{x}{2}+\frac{3}{2}$
b) Point $(6, 15)$ ; $y=8x-33$
c) Two tangent lines: $y=2x-1$ and $y=-2x-1$
d) Slopes with product $-1$ ; vertical normal $x=a$

a)

Point of tangency: $f(3)=9-12+3=0$ , so $(3,0)$ . Slope: $f'(x)=2x-4$ , so $f'(3)=2$ .

Tangent line: $y-0=2(x-3)$ , that is $y=2x-6$ .

Normal line, perpendicular to the tangent at the same point, with slope $-\frac{1}{2}$ :

\displaystyle y=-\frac{1}{2}(x-3)=-\frac{x}{2}+\frac{3}{2}

Check: the product of the slopes is $2\times\left(-\frac{1}{2}\right)=-1$ ✓, the condition for perpendicularity.

b)

Parallel means the same slope: look for $x$ such that $f'(x)=8$ , that is $2x-4=8$ , hence $x=6$ .

Point of tangency: $f(6)=36-24+3=15$ , that is $(6,15)$ .

\displaystyle y-15=8(x-6)\iff y=8x-33

The method is the reverse of a): there we knew the point and looked for the slope, here we know the slope and look for the point. Every tangent line question comes down to one of these two directions.

c)

Pitfall: $P(0,-1)$ is not on the curve, so you cannot compute the derivative there.

Use the unknown point of tangency as the parameter: $(a,a^{2})$ . The slope there is $2a$ , hence the tangent line:

\displaystyle y-a^{2}=2a(x-a)

Require it to pass through $P(0,-1)$ :

\begin{aligned}-1-a^{2} &= 2a(0-a) = -2a^{2} \\ a^{2} &= 1 \quad\text{so}\quad a=\pm 1\end{aligned}

There are therefore two tangent lines. For $a=1$ : point of tangency $(1,1)$ , slope 2, $y=2x-1$ . For $a=-1$ : point of tangency $(-1,1)$ , slope $-2$ , $y=-2x-1$ .

Check: both lines do pass through $(0,-1)$ ✓. From a point below a parabola, you see exactly two tangent lines, symmetric about the axis.

d)

Two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$ . Since the tangent line has slope $f'(a)$ , the slope $m$ of the normal line satisfies:

\displaystyle m\cdot f'(a)=-1\iff m=-\frac{1}{f'(a)}

If $f'(a)=0$ , the tangent line is horizontal and the formula loses its meaning (division by zero): the normal line is then vertical, with equation $x=a$ .

A vertical line has no slope, so no formula in $m$ could describe it: this case is handled separately. It is the same phenomenon as in part d) of exercise 7 on vertical tangent lines, with the roles of the tangent and the normal swapped.

Exercise 6: Transcendental functions

The exponential, the logarithm and trigonometry complete the toolbox; combined with the earlier rules, they cover almost every function in the course.

a) Differentiate $f(x)=e^{2x}$ , $g(x)=\ln(x^{2}+1)$ and $h(x)=xe^{x}$ .
b) Differentiate $u(x)=\sin x\cos x$ in two ways (with the product rule, then after simplifying with the double-angle formula) and check that the results agree.
c) Differentiate $v(x)=e^{x}\ln x$ and $w(x)=\frac{\ln x}{x}$ . For $w$ , find the point where the tangent line is horizontal.
d) Compute the derivative of $y=x^{x}$ (for $x>0$ ) by LOGARITHMIC DIFFERENTIATION. Why does neither the power rule nor the exponential rule apply directly?

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Answers

a) $2e^{2x}$ ; $\frac{2x}{x^{2}+1}$ ; $e^{x}(1+x)$
b) $u'(x)=\cos 2x$
c) $v'=e^{x}\left(\ln x+\frac{1}{x}\right)$ ; $w'=\frac{1-\ln x}{x^{2}}$ equals zero at $x=e$
d) $y'=x^{x}(\ln x+1)$

a)

$f'(x)=e^{2x}\cdot 2=2e^{2x}$ (chain rule: the inside $2x$ has derivative 2).

\displaystyle g'(x)=\frac{1}{x^{2}+1}\cdot 2x=\frac{2x}{x^{2}+1}

\displaystyle h'(x)=(1)e^{x}+xe^{x}=e^{x}(1+x)

For $h$ , a product: the exponential can always be factored out.

b)

With the product rule:

\displaystyle u'=\cos x\cos x+\sin x(-\sin x)=\cos^{2}x-\sin^{2}x

After simplifying, $u(x)=\frac{1}{2}\sin 2x$ :

\displaystyle u'=\frac{1}{2}\cos(2x)\cdot 2=\cos 2x

The two results agree, since $\cos 2x=\cos^{2}x-\sin^{2}x$ : that is exactly the double-angle identity.

Numerical check at $x=1$ : $\cos^{2}1-\sin^{2}1\approx 0.2919-0.7081=-0.4161$ and $\cos 2\approx-0.4161$ ✓.

Simplifying before differentiating often gives the most readable form.

c)

\displaystyle v'=e^{x}\ln x+e^{x}\cdot\frac{1}{x}=e^{x}\left(\ln x+\frac{1}{x}\right)

For $w=\frac{\ln x}{x}$ , a quotient:

\begin{aligned}w' &= \frac{\frac{1}{x}\cdot x-\ln x\cdot 1}{x^{2}} \\ &= \frac{1-\ln x}{x^{2}}\end{aligned}

Horizontal tangent when $w'=0$ , that is $\ln x=1$ , so $x=e$ , where $w(e)=\frac{1}{e}\approx 0.368$ .

This is the maximum of the function, and the fact that it falls exactly at $x=e$ is the reason why $e^{\pi}>\pi^{e}$ : one of the classic comparisons at the end of the chapter.

d)

Neither one applies: the power rule $(x^{n})'=nx^{n-1}$ requires a constant exponent, and $(a^{x})'=a^{x}\ln a$ requires a constant base. Here both the base and the exponent vary.

Method: $\ln y=x\ln x$ , then differentiate both sides with respect to $x$ (the left side with the chain rule, since $y$ depends on $x$ ):

\begin{aligned}\frac{y'}{y} &= \ln x+x\cdot\frac{1}{x} = \ln x+1 \\ y' &= x^{x}(\ln x+1)\end{aligned}

Plausibility check: $y'=0$ for $\ln x=-1$ , that is $x=\frac{1}{e}\approx 0.368$ , and $x^{x}$ does reach its minimum $\approx 0.692$ there: the formula delivers the right critical number.

Exercise 7: Implicit differentiation

The figure shows the circle $x^{2}+y^{2}=25$ and its point $(3,4)$ : no function $y=f(x)$ describes the whole circle, and yet the tangent line at this point exists. Implicit differentiation finds it without ever solving for $y$ .

a) Explain why the circle in the figure is not the graph of a function, and what this prevents you from doing.
b) Differentiate $x^{2}+y^{2}=25$ implicitly, obtain $\frac{dy}{dx}$ , and give the equation of the tangent line at the point $(3,4)$ . Check that this tangent line is perpendicular to the radius.
c) The folium of Descartes has equation $x^{3}+y^{3}=6xy$ . Find $\frac{dy}{dx}$ , then the slope of the tangent line at the point $(3,3)$ .
d) Find $\frac{dy}{dx}$ for the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ , and deduce the points where the tangent line is horizontal and those where it is vertical.

Show the solution

Answers

a) Two points for $x=3$ : not a function
b) $\frac{dy}{dx}=-\frac{x}{y}$ ; $y=-\frac{3}{4}x+\frac{25}{4}$
c) $\frac{2y-x^{2}}{y^{2}-2x}$ ; slope $-1$ at $(3, 3)$
d) Horizontal at $(0, \pm 2)$ ; vertical at $(\pm 3, 0)$

a)

The circle fails the vertical line test: for $x=3$ , there are two points, $(3,4)$ and $(3,-4)$ .

So it is not the graph of a function, and you cannot write $y=f(x)$ globally, only branch by branch, $y=\pm\sqrt{25-x^{2}}$ , at the cost of an arbitrary choice and of square roots that are tedious to differentiate.

Implicit differentiation gets around the obstacle: it differentiates the relation as it stands.

b)

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ :

\displaystyle 2x+2y\frac{dy}{dx}=0\iff\frac{dy}{dx}=-\frac{x}{y}

At $(3,4)$ , the slope is $-\frac{3}{4}$ . Tangent line, the blue line in the figure:

\displaystyle y-4=-\frac{3}{4}(x-3)\iff y=-\frac{3}{4}x+\frac{25}{4}

Perpendicularity: the radius from $(0,0)$ to $(3,4)$ has slope $\frac{4}{3}$ , and $\frac{4}{3}\times\left(-\frac{3}{4}\right)=-1$ ✓. The formula $-\frac{x}{y}$ encodes this property for all the points of the circle at once.

c)

Differentiate term by term, the right-hand side requiring the product rule:

\displaystyle 3x^{2}+3y^{2}\frac{dy}{dx}=6y+6x\frac{dy}{dx}

Collect the terms in $\frac{dy}{dx}$ :

\begin{aligned}\frac{dy}{dx}(3y^{2}-6x) &= 6y-3x^{2} \\ \frac{dy}{dx} &= \frac{2y-x^{2}}{y^{2}-2x}\end{aligned}

At $(3,3)$ , which does satisfy $27+27=54=6\cdot 9$ : $\frac{6-9}{9-6}=-1$ .

The slope $-1$ reflects the symmetry of the folium about the line $y=x$ , on which $(3,3)$ lies.

d)

\displaystyle \frac{2x}{9}+\frac{2y}{4}\frac{dy}{dx}=0\iff\frac{dy}{dx}=-\frac{4x}{9y}

Horizontal tangent when the slope is zero, that is $x=0$ : the points $(0,2)$ and $(0,-2)$ , the vertices on the minor axis.

Vertical tangent when the slope does not exist, that is $y=0$ : the points $(3,0)$ and $(-3,0)$ , the vertices on the major axis.

Where the numerator vanishes, the curve is flat; where the denominator vanishes, it is vertical.

Exercise 8: Differentiability: the three ways to lose a derivative

The figure shows $y=|x|$ (the corner) and $y=x^{2/3}$ (the cusp): two functions that are perfectly continuous at 0, and yet have no derivative at that point.

a) Show that $f(x)=|x|$ is continuous at 0 but not differentiable there, by computing the two one-sided derivatives.
b) State the theorem linking differentiability and continuity, specify the direction of the implication, and explain why a) is a counterexample to the converse.
c) Study the differentiability of $g(x)=x^{1/3}$ at 0 (vertical tangent) and of $h(x)=x^{2/3}$ at 0 (cusp). How do these two failures differ from each other, and from the corner of $|x|$ ?
d) Find $a$ and $b$ so that the following function is DIFFERENTIABLE everywhere: $f(x)=x^{2}$ if $x\le 1$ , and $f(x)=ax+b$ if $x>1$ . How many conditions are needed, and why is continuity alone not enough?

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Answers

a) One-sided derivatives $-1$ and $1$ : not differentiable
b) Differentiable implies continuous, not the converse
c) $x^{1/3}$ : vertical tangent; $x^{2/3}$ : cusp
d) $a=2$ , $b=-1$ : two conditions

a)

Continuity: $\lim_{x\to 0^{-}}|x|=0=\lim_{x\to 0^{+}}|x|=|0|$ , the three conditions are met.

One-sided derivatives:

\begin{aligned}\lim_{h\to 0^{+}}\frac{|h|-0}{h} &= \lim\frac{h}{h} = 1 \\ \lim_{h\to 0^{-}}\frac{|h|-0}{h} &= \lim\frac{-h}{h} = -1\end{aligned}

Both exist but they differ: the two-sided limit does not exist, so $f'(0)$ does not exist.

Geometrically, the graph has a corner: the slope jumps from $-1$ to $+1$ , and no single tangent line can be defined.

b)

Theorem: if $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . The implication goes in that direction only.

Idea of the proof:

\displaystyle f(a+h)-f(a)=\frac{f(a+h)-f(a)}{h}\cdot h\to f'(a)\cdot 0=0

The converse is false, and $|x|$ is the standard counterexample: continuous at 0 without being differentiable there.

Having no hole is not enough; there must be no angle either.

c)

For $g(x)=x^{1/3}$ :

\displaystyle g'(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3]{x^{2}}}\to+\infty

on both sides, since the denominator is always positive. The curve has a vertical tangent at 0: a tangent line can be drawn there, but its slope is infinite, so $g'(0)$ does not exist as a number.

For $h(x)=x^{2/3}$ :

\displaystyle h'(x)=\frac{2}{3}x^{-1/3}

which tends to $+\infty$ on the right and to $-\infty$ on the left: the two branches arrive vertically from opposite directions and form a point, the cusp in the figure.

The corner of $|x|$ has two distinct finite slopes; the cusp has two infinite slopes of opposite signs; the vertical tangent has a single infinite slope. Three pathologies, one symptom: no derivative.

d)

Two conditions: the pieces must meet (continuity) and their slopes must match.

Continuity at 1: $1^{2}=a(1)+b$ , that is $a+b=1$ .

Derivatives at 1: on the left $(x^{2})'=2x$ equals 2, on the right $(ax+b)'=a$ .

\displaystyle a=2,\qquad b=1-2=-1

The line $y=2x-1$ is precisely the tangent line to the parabola at $(1,1)$ : the only smooth junction possible extends the curve in its own direction.

Pitfall: settling for continuity. It glues the pieces together but still allows a corner, exactly as with $|x|$ .

Exercise 9: Inverse functions and their derivatives

An inverse function swaps the $x$ - and $y$ -coordinates, so it also swaps the slopes. A single formula follows, and along the way it gives the derivatives of the inverse sine and the inverse tangent.

a) Establish the formula $\left(f^{-1}\right)'(b)=\frac{1}{f'(a)}$ , where $b=f(a)$ . What geometric interpretation can you give it?
b) Find $\frac{d}{dx}\arcsin x$ by implicit differentiation of $\sin y=x$ , justifying the sign of the square root.
c) In the same way, find $\frac{d}{dx}\arctan x$ from $\tan y=x$ .
d) Let $f(x)=x^{3}+x+1$ . Compute $\left(f^{-1}\right)'(3)$ without finding $f^{-1}$ explicitly. Why is this essential here?

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Answers

a) $\left(f^{-1}\right)'(b)=\frac{1}{f'(a)}$ : reflection in $y=x$
b) $\frac{1}{\sqrt{1-x^{2}}}$ on $(-1,1)$
c) $\frac{1}{1+x^{2}}$ on $\mathbb{R}$
d) $\left(f^{-1}\right)'(3)=\frac{1}{4}$

a)

By definition of the inverse, $f\left(f^{-1}(x)\right)=x$ . Differentiate both sides with the chain rule:

\begin{aligned}f'\left(f^{-1}(x)\right)\cdot\left(f^{-1}\right)'(x) &= 1 \\ \left(f^{-1}\right)'(x) &= \frac{1}{f'\left(f^{-1}(x)\right)}\end{aligned}

At $x=b$ , with $f^{-1}(b)=a$ : $\left(f^{-1}\right)'(b)=\frac{1}{f'(a)}$ ✓.

Interpretation: the graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y=x$ ; the reflection swaps $\Delta x$ and $\Delta y$ , so it turns a slope into its reciprocal.

Consequence: if $f'(a)=0$ , the horizontal tangent becomes vertical, and $f^{-1}$ is not differentiable at $b$ .

b)

Let $y=\arcsin x$ , that is $\sin y=x$ with $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ . Differentiate:

\displaystyle \cos y\cdot\frac{dy}{dx}=1\iff\frac{dy}{dx}=\frac{1}{\cos y}

From $\sin^{2}y+\cos^{2}y=1$ you get $\cos y=\pm\sqrt{1-x^{2}}$ . On $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , the cosine is positive or zero: keep the positive root.

\displaystyle \frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^{2}}},\quad -1<x<1

At the endpoints $x=\pm 1$ , the derivative blows up: these are the vertical tangents of the inverse sine.

c)

Let $y=\arctan x$ , that is $\tan y=x$ with $y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ . Differentiate:

\displaystyle \sec^{2}y\cdot\frac{dy}{dx}=1\iff\frac{dy}{dx}=\frac{1}{\sec^{2}y}

The identity $1+\tan^{2}y=\sec^{2}y$ gives $\sec^{2}y=1+x^{2}$ directly:

\displaystyle \frac{d}{dx}\arctan x=\frac{1}{1+x^{2}},\quad x\in\mathbb{R}

No sign discussion here, since $\sec^{2}y$ is a square. It is this derivative that will make the inverse tangent an antiderivative of $\frac{1}{1+x^{2}}$ in the integral calculus course.

d)

Look for $a$ such that $f(a)=3$ : by trial, $f(1)=1+1+1=3$ ✓, so $f^{-1}(3)=1$ .

Then $f'(x)=3x^{2}+1$ , so $f'(1)=4$ , and:

\displaystyle \left(f^{-1}\right)'(3)=\frac{1}{f'(1)}=\frac{1}{4}

This is essential because $f^{-1}$ has no explicit formula: inverting $y=x^{3}+x+1$ would require solving a general cubic. The formula needs only the value of $a$ , found by inspection, and the derivative of $f$ , which is elementary: you compute the slope of a function you cannot write down.

Incidentally, $f'(x)=3x^{2}+1>0$ everywhere: $f$ is strictly increasing, so it really is invertible. This check is part of the answer.

Exercise 10: Proofs: the derivative of sine and the quotient rule

The formulas on the formula sheet do not fall from the sky: each one is proved from the definition and from the limits of the previous chapter. The exam regularly asks for one.

a) Prove that $(\sin x)'=\cos x$ from the DEFINITION of the derivative. Which limits from the chapter on limits are essential?
b) Prove the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ from the product rule and the chain rule.
c) Prove that $(x^{n})'=nx^{n-1}$ still holds for a NEGATIVE integer $n$ , by setting $n=-m$ with $m>0$ .
d) Reconstruct the chain of dependencies: which results must have been established, and in which order, for the proof in part a) to be legitimate?

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Answers

a) $(\sin x)'=\cos x$ thanks to $\frac{\sin h}{h}\to 1$ and $\frac{\cos h-1}{h}\to 0$
b) $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$
c) $(x^{-m})'=-mx^{-m-1}$
d) Limits, trigonometric limits, definition, addition formula; not L'Hôpital's rule

a)

From the definition, with the addition formula $\sin(x+h)=\sin x\cos h+\cos x\sin h$ :

\begin{aligned}&\frac{\sin(x+h)-\sin x}{h} \\ &\quad = \frac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\ &\quad = \sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\end{aligned}

The two essential limits, established in the chapter on limits (the second one with the conjugate):

\displaystyle \lim_{h\to 0}\frac{\sin h}{h}=1\qquad\lim_{h\to 0}\frac{\cos h-1}{h}=0

\displaystyle (\sin x)'=\sin x\cdot 0+\cos x\cdot 1=\cos x

$x$ is constant in this computation, and it is $h$ that tends to 0: $\sin x$ and $\cos x$ come out of the limits as constants.

b)

Write the quotient as a product, $\frac{u}{v}=u\cdot v^{-1}$ , then use the product and chain rules, with $\left(v^{-1}\right)'=-v^{-2}v'$ :

\begin{aligned}\left(\frac{u}{v}\right)' &= u'v^{-1}+u\left(v^{-1}\right)' \\ &= \frac{u'}{v}-\frac{uv'}{v^{2}} \\ &= \frac{u'v-uv'}{v^{2}}\end{aligned}

The proof explains why the order of the terms cannot be swapped: the minus sign comes from the derivative of $v^{-1}$ , so it applies to the term that contains $v'$ . It is the best way never to swap the two terms on an exam again.

c)

Let $n=-m$ with $m>0$ an integer, so that $x^{n}=\frac{1}{x^{m}}$ . Quotient rule with $u=1$ , $u'=0$ , $v=x^{m}$ , $v'=mx^{m-1}$ :

\begin{aligned}\left(x^{-m}\right)' &= \frac{0\cdot x^{m}-1\cdot mx^{m-1}}{\left(x^{m}\right)^{2}} \\ &= \frac{-mx^{m-1}}{x^{2m}} = -mx^{-m-1}\end{aligned}

Going back to $n=-m$ : $-mx^{-m-1}=nx^{n-1}$ ✓. The rule therefore holds for every integer.

Check: for $n=-2$ , the formula predicts $\left(x^{-2}\right)'=-2x^{-3}$ , which agrees with the computation of $f''$ in exercise 4 c).

d)

The chain, from the most fundamental to the most derived:

1. The definition of the limit and its algebraic properties.

2. The limits $\lim\frac{\sin h}{h}=1$ and $\lim\frac{\cos h-1}{h}=0$ , which require the Squeeze Theorem and measuring angles in radians (in degrees, the first one would equal $\frac{\pi}{180}$ and every trigonometric derivative would carry this stray factor).

3. The definition of the derivative as the limit of the difference quotient.

4. The addition formula for $\sin(a+b)$ , which belongs to trigonometry, not to calculus.

Only then is the proof in a) legitimate.

Pitfall: using L'Hôpital's rule to establish $\lim\frac{\sin h}{h}=1$ . The reasoning is circular, since L'Hôpital's rule assumes the derivative of sine is known, which is precisely what is being proved. It is the most frequent rigour trap in the course.

Exercise 11: The rules from a table of values

The functions $f$ and $g$ are known only through the following table, which also gives their derivatives. Both are differentiable on $\mathbb{R}$ .

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$1$	$3$	$-2$	$2$	$4$
$2$	$3$	$5$	$3$	$-1$
$3$	$4$	$2$	$1$	$3$

a) Compute $(f+3g)'(2)$ .
b) Compute $(fg)'(1)$ .
c) Compute $\left(\frac{f}{g}\right)'(2)$ .
d) Compute $(f\circ g)'(1)$ and $(g\circ f)'(2)$ .
e) Compute the derivative of $f^{2}$ at $1$ and that of $\sqrt{f}$ at $3$ .

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Answers

a) $2$
b) $8$
c) $2$
d) $20$ and $15$
e) $-12$ and $\frac{1}{2}$

a)

The derivative of a sum is the sum of the derivatives, and a constant factor comes out:

\displaystyle (f+3g)'(2)=f'(2)+3g'(2)=5+3\times(-1)=2

b)

\displaystyle (fg)'(1)=f'(1)g(1)+f(1)g'(1)=(-2)(2)+(3)(4)=8

Pitfall: the product of the derivatives, $f'(1)g'(1)=-8$ , would even give the wrong sign.

c)

\begin{aligned}\left(\frac{f}{g}\right)'(2) &= \frac{f'(2)g(2)-f(2)g'(2)}{g(2)^{2}} \\ &= \frac{5\times 3-3\times(-1)}{9} = \frac{18}{9} = 2\end{aligned}

Pitfall: the minus sign applies to the term that contains $g'$ ; reversing the order would give $-2$ .

d)

$(f\circ g)'(1)=f'\left(g(1)\right)\cdot g'(1)$ . First read $g(1)=2$ , then $f'(2)=5$ :

\displaystyle (f\circ g)'(1)=5\times 4=20

Similarly:

\displaystyle (g\circ f)'(2)=g'\left(f(2)\right)\cdot f'(2)=g'(3)\times 5=3\times 5=15

The two orders give two different numbers, just as the two composite functions are two different functions.

Pitfall: $f'(1)\cdot g'(1)=-8$ . The outside derivative is evaluated at $g(1)$ , not at $1$ .

e)

$f^{2}$ is the composition of $u\mapsto u^{2}$ with $f$ :

\displaystyle (f^{2})'(1)=2f(1)f'(1)=2\times 3\times(-2)=-12

\displaystyle (\sqrt{f})'(3)=\frac{f'(3)}{2\sqrt{f(3)}}=\frac{2}{2\times 2}=\frac{1}{2}

In both cases, the chain rule needed the value of $f$ at the point, not just its derivative: that is the row of the table people forget.

Exercise 12: The common tangent to two parabolas

Consider $f(x)=x^{2}$ and $g(x)=-x^{2}+2x-5$ , whose graphs are shown in the figure. We are looking for the lines that are tangent to BOTH curves at once.

a) Write the equation of the tangent line to the graph of $f$ at the point with $x$ -coordinate $a$ .
b) Write the equation of the tangent line to the graph of $g$ at the point with $x$ -coordinate $b$ .
c) Write the two conditions for these two lines to coincide, and solve the system.
d) Give the equations of the common tangent lines.
e) Give the points of tangency on the graph of $g$ and check one of them.

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Answers

a) $y=2ax-a^{2}$
b) $y=(2-2b)x+b^{2}-5$
c) $a=2$ , $b=-1$ or $a=-1$ , $b=2$
d) $y=4x-4$ and $y=-2x-1$
e) $(-1, -8)$ and $(2, -5)$

a)

$f(a)=a^{2}$ and $f'(a)=2a$ :

\displaystyle y=a^{2}+2a(x-a)\iff y=2ax-a^{2}

b)

$g(b)=-b^{2}+2b-5$ and $g'(b)=-2b+2$ :

\begin{aligned}y &= -b^{2}+2b-5+(-2b+2)(x-b) \\ &= (2-2b)x+b^{2}-5\end{aligned}

because the constant term is $-b^{2}+2b-5+2b^{2}-2b=b^{2}-5$ .

c)

Two lines coincide when they have the same slope and the same $y$ -intercept:

\displaystyle 2a=2-2b\qquad -a^{2}=b^{2}-5

The first gives $b=1-a$ . Substituting:

\begin{aligned}-a^{2} &= (1-a)^{2}-5 = a^{2}-2a-4 \\ 0 &= a^{2}-a-2 = (a-2)(a+1)\end{aligned}

So $a=2$ , $b=-1$ , or $a=-1$ , $b=2$ .

Pitfall: matching only the slopes. You get infinitely many pairs of parallel tangent lines, not a common tangent.

d)

For $a=2$ : $y=4x-4$ . For $a=-1$ : $y=-2x-1$ .

Two common tangent lines, consistent with the figure: one rises, the other falls, and the two parabolas, opening in opposite directions, do not intersect.

e)

For $y=4x-4$ , tangency on $g$ at $b=-1$ : $g(-1)=-1-2-5=-8$ , point $(-1, -8)$ .

Check: $g'(-1)=2+2=4$ , the slope is right, and $4\times(-1)-4=-8$ , the point is on the line ✓.

For $y=-2x-1$ , tangency at $b=2$ : $g(2)=-5$ , point $(2, -5)$ , with $g'(2)=-2$ and $-2\times 2-1=-5$ ✓.

A tangent line is checked by two tests, the slope and whether the point lies on the line: one without the other proves nothing.

Exercise 13: The numerical derivative: estimating a slope from measurements

A tank is being filled, and the volume of water it contains is recorded every minute. The table gives the volume $V$ , in litres, as a function of the time $t$ , in minutes.

$t$ (min)	$1$	$2$	$3$	$4$	$5$
$V$ (L)	$537$	$568$	$593$	$612$	$625$

a) Compute the average rate of change of the volume between $t=1$ and $t=5$ .
b) Estimate $V'(3)$ with the FORWARD difference $\frac{V(4)-V(3)}{1}$ , then with the BACKWARD difference $\frac{V(3)-V(2)}{1}$ .
c) Estimate $V'(3)$ with the CENTRAL difference $\frac{V(4)-V(2)}{2}$ . Why is it generally better?
d) The measurements actually follow $V(t)=500+40t-3t^{2}$ . Compute $V'(3)$ exactly, then show that the central difference is EXACT for this model, at every point and for every step size.
e) At what time does the filling rate become zero? Was the table enough to tell?

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Answers

a) $22$ L/min
b) $19$ and $25$ L/min
c) $22$ L/min
d) $V'(3)=22$ : the central difference is exact
e) $t\approx 6.67$ min; the table did not show it

a)

$\frac{V(5)-V(1)}{5-1}=\frac{625-537}{4}=22$ L/min

This is an average rate: it says what happened over four minutes, not at a precise instant.

b)

Forward: $\frac{612-593}{1}=19$ L/min. Backward: $\frac{593-568}{1}=25$ L/min.

The two estimates bracket the true value but each one is off by $3$ : the curve bends, and a secant line taken on one side only inherits that curvature.

c)

$\frac{612-568}{2}=22$ L/min, the average of the two previous ones.

A central secant line offsets the curvature error on one side with the error on the other: the error of the forward difference is proportional to the step size, that of the central difference to the square of the step size. This is the formula used by calculators and spreadsheets.

d)

$V'(t)=40-6t$ , so $V'(3)=22$ L/min, exactly the central estimate. In general:

\begin{aligned}&V(a+k)-V(a-k) \\ &\quad = 40(2k)-3\left[(a+k)^{2}-(a-k)^{2}\right] \\ &\quad = 80k-12ak\end{aligned}

Dividing by $2k$ : $40-6a=V'(a)$ , whatever $a$ and $k$ are.

The terms in $k^{2}$ cancelled by symmetry: for a polynomial of degree $2$ , the central difference makes no error at all. The average rate in a), between $1$ and $5$ , is in fact the central difference at $3$ with step $2$ .

e)

$V'(t)=0$ gives $t=\frac{40}{6}\approx 6.67$ min, where the volume peaks at $V\approx 633.3$ L before decreasing: the tank starts to leak faster than it fills.

The table, which stops at $t=5$ , did not show it: all the rates in it are positive. A numerical estimate says nothing beyond the measurements; only the model can predict.

Exercise 14: Problem: the path of a ball and the angle of the tangent

A ball is thrown from a height of $2$ m. Its path, in a vertical plane, follows $y=-0.05x^{2}+0.9x+2$ , where $x$ is the horizontal distance from the thrower and $y$ the height, both in metres. The angle of a path with the horizontal at a point is the angle $\theta$ such that $\tan\theta=y'(x)$ . The figure below shows the situation; it is not to scale.

a) Compute $y'(x)$ and the launch angle, to the nearest tenth of a degree.
b) Where does the ball reach its maximum height, and what is that height?
c) A net $3$ m high is set up $16$ m from the thrower. Does the ball clear it, and by how much?
d) Where does the ball hit the ground, and at what angle?
e) At which point is the path parallel to the segment joining the launch point to the point of impact?

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Answers

a) $y'(x)=-0.1x+0.9$ ; $42.0°$
b) $x=9$ m, $6.05$ m
c) $3.6$ m: margin $0.6$ m
d) $x=20$ m, $47.7°$ below the horizontal
e) $x=10$ m

a)

$y'(x)=-0.1x+0.9$ , so $y'(0)=0.9$ and:

\displaystyle \theta=\arctan 0.9\approx 42.0°

The derivative gives a slope, a dimensionless number; the angle follows from the inverse tangent.

b)

At the top, the tangent line is horizontal: $y'(x)=0$ gives $x=9$ m.

$y(9)=-4.05+8.1+2=6.05$ m

c)

$y(16)=-12.8+14.4+2=3.6$ m

The ball clears the net with a margin of $0.6$ m.

d)

\begin{aligned}-0.05x^{2}+0.9x+2 &= 0 \\ x^{2}-18x-40 &= 0 \\ (x-20)(x+2) &= 0\end{aligned}

The negative root is outside the domain: impact at $x=20$ m. There:

\displaystyle y'(20)=-2+0.9=-1.1\qquad\theta=\arctan(-1.1)\approx-47.7°

The ball comes down at $47.7°$ below the horizontal, steeper than at launch, because it falls from higher up than where it started.

e)

The segment joins $(0, 2)$ to $(20, 0)$ : slope $\frac{0-2}{20}=-0.1$ .

$-0.1x+0.9=-0.1\iff x=10$ m, at the point $(10, 6)$ .

Such a point always exists for a differentiable curve, which is what the Mean Value Theorem will guarantee; for a parabola, it falls exactly at the midpoint of the interval.

Exercise 15: Problem: joining two roads without a jolt

A horizontal road, at height $0$ for $x\le 0$ , must join a second horizontal road, at height $2$ m for $x\ge 40$ , all distances being in metres. We are looking for a transition curve $p(x)=ax^{3}+bx^{2}+cx+d$ on $[0,40]$ that joins WITHOUT A JOLT: the same height and the same slope as the road at each end. The figure in the solution shows the resulting profile. The figure below shows the situation.

a) Write the four conditions for the junction. Why can a parabola $bx^{2}+cx+d$ not work?
b) Find $a$ , $b$ , $c$ and $d$ .
c) Compute the height of the road halfway along.
d) Where is the slope steepest, and what is it as a percentage?
e) We now want to climb $3$ m instead of $2$ , with a maximum grade of $10$ %. What is the minimum length of the transition?

Show the solution

Answers

a) $p(0)=p'(0)=0$ , $p(40)=2$ , $p'(40)=0$ : four conditions
b) $a=-6.25\times 10^{-5}$ , $b=3.75\times 10^{-3}$ , $c=d=0$
c) $1$ m
d) At $x=20$ m: $7.5$ %
e) $L\ge 45$ m

a)

Four conditions: $p(0)=0$ and $p'(0)=0$ at the start, $p(40)=2$ and $p'(40)=0$ at the end.

A parabola has only three coefficients for four conditions. Concretely, $p(0)=0$ and $p'(0)=0$ force $d=c=0$ , so $p(x)=bx^{2}$ ; then $p'(40)=80b=0$ forces $b=0$ , and $p(40)=0\neq 2$ .

Degree $3$ is needed.

b)

$p'(x)=3ax^{2}+2bx+c$ . The first two conditions give $d=0$ and $c=0$ . The other two:

\begin{aligned}64000a+1600b &= 2 \\ 4800a+80b &= 0\end{aligned}

The second gives $b=-60a$ ; substituting, $64000a-96000a=2$ :

\displaystyle a=-\frac{1}{16000}=-6.25\times 10^{-5}\qquad b=3.75\times 10^{-3}

\displaystyle p(x)=0.00375x^{2}-0.0000625x^{3}

Check: $p(40)=6-4=2$ ✓ and $p'(40)=0.3-0.3=0$ ✓.

c)

$p(20)=1.5-0.5=1$ m

Exactly half of the climb: the profile is symmetric about its midpoint.

d)

The slope $p'(x)=0.0075x-0.0001875x^{2}$ is greatest where its own derivative is zero:

$p''(x)=0.0075-0.000375x=0\iff x=20$ m

There, $p'(20)=0.15-0.075=0.075$ , that is a grade of $7.5$ %.

This is the inflection point of the profile, where the road stops steepening and starts to level off.

e)

With a climb $H$ over a length $L$ , the same computation gives:

\displaystyle p(x)=H\left[3\left(\frac{x}{L}\right)^{2}-2\left(\frac{x}{L}\right)^{3}\right]

with maximum slope $\frac{3H}{2L}$ at the midpoint; this gives back $\frac{3\times 2}{80}=0.075$ . For $H=3$ and $L=40$ , it would rise to $11.25$ %.

To stay below $10$ %:

$\frac{3\times 3}{2L}\le 0.10\iff L\ge 45$ m

Corrected exercises: the derivative and its rules (201-SN2-RE)

Course recap

Calculations: the derivative from the difference quotient

Exercise 0.1: The derivative at a point from the difference quotient

Exercise 0.2: The derivative function from the difference quotient

Part A: Defining and computing the derivative (/50)

Exercise 1: From the secant line to the tangent line: the definition

Exercise 2: The rules: power, product, quotient

Exercise 3: The chain rule: differentiating compositions

Exercise 4: Higher-order derivatives

Exercise 5: Tangent and normal lines

Part B: Transcendental functions, inverses and proofs (/50)

Exercise 6: Transcendental functions

Exercise 7: Implicit differentiation

Exercise 8: Differentiability: the three ways to lose a derivative

Exercise 9: Inverse functions and their derivatives

Exercise 10: Proofs: the derivative of sine and the quotient rule

Part C: the classics (/50)

Exercise 11: The rules from a table of values

Exercise 12: The common tangent to two parabolas

Exercise 13: The numerical derivative: estimating a slope from measurements

Exercise 14: Problem: the path of a ball and the angle of the tangent

Exercise 15: Problem: joining two roads without a jolt

See also

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