Exercise 0.1: Twelve derivatives to compute directly

For each function, compute the derivative with the differentiation formulas, without simplifying more than necessary, then evaluate it at the given point. The notations tg, cotg and cosec found in some French-language textbooks stand for $\tan$ , $\cot$ and $\csc$ .

One method for all twelve lines: spot the LAST operation (sum, product, quotient, power or composite function), apply the rule that goes with it, then multiply by the derivative of the inside at every layer. The calculator, in radian mode, is only for the final value.

a) $f(x)=\cot(5x^{2}+x)$ , then $f'(0.3)$ .
b) $g(t)=t^{4}\csc(2t)$ , then $g'(1)$ .
c) $h(x)=\sec^{2}(3x)$ , then $h'(0.2)$ .
d) $y=\sqrt{\csc(x^{2})}$ , then $y'(1)$ .
e) $f(x)=e^{\cos(3x)}$ , then $f'(0.5)$ .
f) $g(x)=5^{\tan x}$ , then $g'\left(\frac{\pi}{4}\right)$ .
g) $h(t)=\ln(\csc t)+\csc(\ln t)$ , then $h'(2)$ .
h) $s(x)=\sin(x^{3})+\sin^{3}x$ , then $s'(1)$ .
i) $f(x)=\arctan(\sqrt{x})$ , then $f'(3)$ .
j) $g(x)=\frac{\arccos(2x)}{x^{2}+1}$ , then $g'(0.25)$ .
k) $h(x)=e^{2x}\cos(x^{2})$ , then $h'(1)$ .
l) $y=\frac{1+\sec(2x)}{1-\tan(2x)}$ , then $y'(0.1)$ .

Type your answers, the page tells you right or wrong 0/12

a)

f'(0.3)

b)

g'(1)

c)

h'(0.2)

d)

y'(1)

e)

f'(0.5)

f)

g'\left(\frac{\pi}{4}\right)

g)

h'(2)

h)

s'(1)

i)

f'(3)

j)

g'(0.25)

k)

h'(1)

l)

y'(0.1)

Show the solution

Answers

a) $f'(x)=-(10x+1)\csc^{2}(5x^{2}+x)$ ; $f'(0.3)\approx-8.609$
b) $g'(t)=2t^{3}\csc(2t)\left(2-t\cot(2t)\right)$ ; $g'(1)\approx 5.406$
c) $h'(x)=6\sec^{2}(3x)\tan(3x)$ ; $h'(0.2)\approx 6.026$
d) $y'=-x\cot(x^{2})\sqrt{\csc(x^{2})}$ ; $y'(1)\approx-0.700$
e) $f'(x)=-3\sin(3x)\,e^{\cos(3x)}$ ; $f'(0.5)\approx-3.212$
f) $g'(x)=5^{\tan x}\ln 5\sec^{2}x$ ; $g'\left(\frac{\pi}{4}\right)=10\ln 5\approx 16.094$
g) $h'(t)=-\cot t-\frac{\csc(\ln t)\cot(\ln t)}{t}$ ; $h'(2)\approx-0.484$
h) $s'(x)=3x^{2}\cos(x^{3})+3\sin^{2}x\cos x$ ; $s'(1)\approx 2.769$
i) $f'(x)=\frac{1}{2\sqrt{x}(1+x)}$ ; $f'(3)=\frac{\sqrt{3}}{24}\approx 0.0722$
j) $g'(x)=\frac{\frac{-2(x^{2}+1)}{\sqrt{1-4x^{2}}}-2x\arccos(2x)}{(x^{2}+1)^{2}}$ ; $g'(0.25)\approx-2.637$
k) $h'(x)=2e^{2x}\left(\cos(x^{2})-x\sin(x^{2})\right)$ ; $h'(1)\approx-4.451$
l) $y'=\frac{2\sec(2x)\tan(2x)(1-\tan(2x))+2\sec^{2}(2x)(1+\sec(2x))}{(1-\tan(2x))^{2}}$ ; $y'(0.1)\approx 7.137$

The move that decides everything happens BEFORE differentiating: name the last operation. $t^{4}\csc(2t)$ is a product, $\sec^{2}(3x)$ a power, $e^{\cos(3x)}$ a composite exponential, $\frac{\arccos(2x)}{x^{2}+1}$ a quotient. Once the rule is chosen, count the layers: each one leaves its inner derivative as a factor.

a)

Composite function, outer function $\cot$ : $f'(x)=-\csc^{2}(5x^{2}+x)\cdot(10x+1)$ . At $x=0.3$ , the argument equals $0.75$ and $10x+1=4$ , so $f'(0.3)=-\frac{4}{\sin^{2}(0.75)}\approx-8.609$ . Pitfall: forgetting the minus sign of the cotangent, which gives $+8.609$ .

b)

Product first: $g'(t)=4t^{3}\csc(2t)+t^{4}\cdot\left(-\csc(2t)\cot(2t)\right)\cdot 2=2t^{3}\csc(2t)\left(2-t\cot(2t)\right)$ . At $t=1$ : $g'(1)=2\csc 2\,(2-\cot 2)\approx 2\times 1.0998\times 2.4577\approx 5.406$ . Careful, $\cot 2$ is NEGATIVE ( $2$ radians is more than $\frac{\pi}{2}$ ), hence the $2.4577$ .

c)

Power first, $h(x)=\left(\sec(3x)\right)^{2}$ : $h'(x)=2\sec(3x)\cdot\sec(3x)\tan(3x)\cdot 3=6\sec^{2}(3x)\tan(3x)$ . At $x=0.2$ : $h'(0.2)=\frac{6\tan(0.6)}{\cos^{2}(0.6)}\approx 6.026$ . Three layers, three factors: the $2$ of the power, the derivative of the secant, the $3$ .

d)

The radical becomes a power: $y=\left(\csc(x^{2})\right)^{\frac{1}{2}}$ . So $y'=\frac{1}{2}\left(\csc(x^{2})\right)^{-\frac{1}{2}}\cdot\left(-\csc(x^{2})\cot(x^{2})\right)\cdot 2x=-x\cot(x^{2})\sqrt{\csc(x^{2})}$ , since $\frac{\csc u}{\sqrt{\csc u}}=\sqrt{\csc u}$ . At $x=1$ : $y'(1)=-\cot 1\sqrt{\csc 1}\approx-0.700$ .

e)

$f'(x)=e^{\cos(3x)}\cdot\left(-\sin(3x)\right)\cdot 3=-3\sin(3x)\,e^{\cos(3x)}$ . The exponential is copied as it is; it is its exponent that gets differentiated. At $x=0.5$ : $f'(0.5)=-3\sin(1.5)\,e^{\cos(1.5)}\approx-3.212$ .

f)

Constant base, variable exponent: $(a^{u})'=a^{u}\ln a\cdot u'$ , so $g'(x)=5^{\tan x}\ln 5\cdot\sec^{2}x$ . At $x=\frac{\pi}{4}$ , $\tan x=1$ and $\sec^{2}x=2$ : $g'\left(\frac{\pi}{4}\right)=10\ln 5\approx 16.094$ . Pitfall: the power rule, $\tan x\cdot 5^{\tan x-1}$ , which only applies when the exponent is constant.

g)

Two terms, two compositions, in opposite orders. $\left(\ln(\csc t)\right)'=\frac{-\csc t\cot t}{\csc t}=-\cot t$ : the logarithm first, then the simplification. $\left(\csc(\ln t)\right)'=-\csc(\ln t)\cot(\ln t)\cdot\frac{1}{t}$ : the cosecant first, then the derivative of $\ln t$ . So $h'(t)=-\cot t-\frac{\csc(\ln t)\cot(\ln t)}{t}$ , and $h'(2)\approx 0.458-0.942=-0.484$ .

h)

$\sin(x^{3})$ : the power is INSIDE, hence $\cos(x^{3})\cdot 3x^{2}$ . $\sin^{3}x=(\sin x)^{3}$ : the power is OUTSIDE, hence $3\sin^{2}x\cos x$ . So $s'(x)=3x^{2}\cos(x^{3})+3\sin^{2}x\cos x$ , and $s'(1)=3\cos 1\left(1+\sin^{2}1\right)\approx 2.769$ .

i)

$f'(x)=\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}(1+x)}$ , for $x>0$ . At $x=3$ : $f'(3)=\frac{1}{8\sqrt{3}}=\frac{\sqrt{3}}{24}\approx 0.0722$ . The square of $\sqrt{x}$ is $x$ : that is what makes the final form so short.

j)

Quotient, with $u=\arccos(2x)$ , $u'=\frac{-2}{\sqrt{1-4x^{2}}}$ , $v=x^{2}+1$ , $v'=2x$ : $g'(x)=\frac{\frac{-2}{\sqrt{1-4x^{2}}}(x^{2}+1)-2x\arccos(2x)}{(x^{2}+1)^{2}}$ , for $|x|<\frac{1}{2}$ . At $x=0.25$ : $u'=\frac{-2}{\sqrt{0.75}}\approx-2.309$ and $\arccos(0.5)=\frac{\pi}{3}$ , so $g'(0.25)=\frac{-2.309\times 1.0625-0.5\times\frac{\pi}{3}}{1.0625^{2}}\approx-2.637$ . Pitfall: the minus sign in the derivative of $\arccos$ .

k)

Product: $h'(x)=2e^{2x}\cos(x^{2})+e^{2x}\cdot\left(-\sin(x^{2})\right)\cdot 2x=2e^{2x}\left(\cos(x^{2})-x\sin(x^{2})\right)$ . At $x=1$ : $h'(1)=2e^{2}(\cos 1-\sin 1)\approx-4.451$ .

l)

Quotient: the numerator $1+\sec(2x)$ has derivative $2\sec(2x)\tan(2x)$ , the denominator $1-\tan(2x)$ has derivative $-2\sec^{2}(2x)$ . So $y'=\frac{2\sec(2x)\tan(2x)\left(1-\tan(2x)\right)+2\sec^{2}(2x)\left(1+\sec(2x)\right)}{\left(1-\tan(2x)\right)^{2}}$ , the minus of the quotient rule and the minus of $v'$ giving a PLUS. At $x=0.1$ : $y'(0.1)\approx 7.137$ .

Overall check: for each line, the rate $\frac{f(a+0.001)-f(a-0.001)}{0.002}$ computed on the calculator lands within a thousandth of the value found. If it does not, look first for a forgotten inner factor, then for the minus sign of a “co” function.

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Exercise 1: The table of the six trigonometric derivatives, rebuilt

You do not learn six derivatives by heart: you learn TWO, those of sine and cosine, and the other four come out of the quotient rule in three lines. This is the thread of the whole set: the chain rule never changes, only the starting table changes.

a) State $(\sin x)'$ and $(\cos x)'$ . Writing $\tan x=\frac{\sin x}{\cos x}$ , prove that $(\tan x)'=\sec^{2}x$ , then show that this expression also equals $1+\tan^{2}x$ .
b) In the same way, establish $(\sec x)'=\sec x\tan x$ and $(\cot x)'=-\csc^{2}x$ .
c) Differentiate $f(x)=\sin(3x)$ , $g(x)=\cos^{2}x$ and $h(x)=\tan(x^{2}+1)$ .
d) The figure shows $y=\tan x$ . At which points is the function undefined, and why can its derivative NEVER be zero?
e) In several French-language Quebec textbooks, $\tan$ is written tg, $\cot$ is written cotg and $\csc$ is written cosec: they are the same functions. Differentiate $f(x)=\mathrm{cotg}(x^{2}-2x)$ and $g(t)=t^{3}\sec(2t)$ , then compute $g'\left(\frac{\pi}{8}\right)$ .
f) Differentiate $h(t)=\csc^{3}(1-4t)$ , counting the minus signs. Then rewrite $k(\theta)=\sqrt[3]{\tan^{2}(5\theta)}$ as a power before differentiating it, compute $k'\left(\frac{\pi}{20}\right)$ , and say what happens at $\theta=0$ .
g) Differentiate $s(x)=\cos(x^{2})+\cos^{2}x$ and compute $s'(1)$ . What distinguishes the way the two terms are written?

Type your answers, the page tells you right or wrong 0/16

a)

(\tan x)'

(\cos x)'

b)

(\sec x)'

(\cot x)'

c)

f'(x)

g'(1)

h'(x)

d)

Minimum value of

(\tan x)'

Is

(\tan x)'

ever zero?

e)

f'(x)

g'\left(\frac{\pi}{8}\right)

f)

h'(t)

k'\left(\frac{\pi}{20}\right)

Is

k

differentiable at

0

?

g)

s'(x)

s'(1)

Show the solution

Answers

a) $(\tan x)'=\sec^{2}x=1+\tan^{2}x$
b) $(\sec x)'=\sec x\tan x$ ; $(\cot x)'=-\csc^{2}x$
c) $3\cos(3x)$ ; $-\sin(2x)$ ; $2x\sec^{2}(x^{2}+1)$
d) Undefined at $\frac{\pi}{2}+k\pi$ ; derivative $\ge 1$
e) $f'(x)=-(2x-2)\csc^{2}(x^{2}-2x)$ ; $g'(t)=t^{2}\sec(2t)\left(3+2t\tan(2t)\right)$ , $g'\left(\frac{\pi}{8}\right)\approx 0.826$
f) $h'(t)=12\csc^{3}(1-4t)\cot(1-4t)$ ; $k'\left(\frac{\pi}{20}\right)=\frac{20}{3}$ ; cusp at $\theta=0$
g) $s'(x)=-2x\sin(x^{2})-\sin(2x)$ ; $s'(1)\approx-2.592$

a)

$(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$ : the minus sign is on the cosine, and that is the only thing to remember. Quotient rule: $(\tan x)'=\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^{2}x}=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=\frac{1}{\cos^{2}x}=\sec^{2}x$ . Splitting the fraction term by term instead: $\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=1+\tan^{2}x$ . The two forms are the same thing, and you choose the one that simplifies what follows: $\sec^{2}x$ when integrating, $1+\tan^{2}x$ when the problem already gives $\tan x$ . Numerical check at $x=0.6$ : $\sec^{2}(0.6)\approx 1.4718$ and $1+\tan^{2}(0.6)\approx 1+0.4718=1.4718$ ✓.

b)

$\sec x=\frac{1}{\cos x}$ , so by the quotient rule $(\sec x)'=\frac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^{2}x}=\frac{\sin x}{\cos^{2}x}=\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}=\sec x\tan x$ . Likewise $\cot x=\frac{\cos x}{\sin x}$ gives $(\cot x)'=\frac{-\sin x\cdot\sin x-\cos x\cos x}{\sin^{2}x}=\frac{-1}{\sin^{2}x}=-\csc^{2}x$ . Remember the symmetry: the three “co” functions (cosine, cotangent, cosecant) all carry a minus sign.

c)

$f'(x)=3\cos(3x)$ : chain rule, the inside $3x$ has derivative 3. $g(x)=(\cos x)^{2}$ so $g'(x)=2\cos x\cdot(-\sin x)=-2\sin x\cos x=-\sin(2x)$ : the compact form is shorter, and it confirms the computation through the double-angle identity. $h'(x)=\sec^{2}(x^{2}+1)\cdot 2x=\frac{2x}{\cos^{2}(x^{2}+1)}$ . The classic trap is to write $g'(x)=-2\sin x$ , forgetting the chain rule: check at $x=1$ , $g'(1)=-\sin 2\approx-0.909$ , whereas $-2\sin 1\approx-1.683$ , almost twice as large.

d)

$\tan x$ is undefined where $\cos x=0$ , that is, at $x=\frac{\pi}{2}+k\pi$ : these are the two vertical asymptotes in the figure. Its derivative $\sec^{2}x=\frac{1}{\cos^{2}x}$ has a square in the denominator, so it is strictly positive wherever it exists, and it is at least 1 (reached when $\cos^{2}x=1$ ). Consequence read on the graph: the tangent function is strictly increasing on each branch and has NO horizontal tangent line, hence no local extremum. A function can be increasing on each piece without being increasing on its whole domain: between two branches, it drops from $+\infty$ to $-\infty$ .

e)

The notation changes nothing: cotg is $\cot$ , and its derivative carries the minus sign of the “co” functions. Chain rule: $f'(x)=-\csc^{2}(x^{2}-2x)\cdot(2x-2)=-(2x-2)\csc^{2}(x^{2}-2x)$ , wherever $\sin(x^{2}-2x)\neq 0$ . For $g$ , it is a PRODUCT before being a composition: $g'(t)=3t^{2}\sec(2t)+t^{3}\cdot\sec(2t)\tan(2t)\cdot 2=t^{2}\sec(2t)\left(3+2t\tan(2t)\right)$ . At $t=\frac{\pi}{8}$ , $2t=\frac{\pi}{4}$ , so $\sec(2t)=\sqrt{2}$ and $\tan(2t)=1$ : $g'\left(\frac{\pi}{8}\right)=\frac{\pi^{2}}{64}\sqrt{2}\left(3+\frac{\pi}{4}\right)\approx 0.826$ . The trap is to differentiate each factor and multiply, $3t^{2}\cdot 2\sec(2t)\tan(2t)$ , which gives about $1.309$ : the derivative of a product is not the product of the derivatives.

f)

Three layers, the power, the cosecant and $1-4t$ : $h'(t)=3\csc^{2}(1-4t)\cdot\left(-\csc(1-4t)\cot(1-4t)\right)\cdot(-4)=12\csc^{3}(1-4t)\cot(1-4t)$ . Two minus signs, that of the cosecant and that of $u'=-4$ , which cancel: write them both down before simplifying, otherwise you keep one of them at random. For $k$ , the radical becomes an exponent: $k(\theta)=\left(\tan(5\theta)\right)^{\frac{2}{3}}$ , so $k'(\theta)=\frac{2}{3}\left(\tan(5\theta)\right)^{-\frac{1}{3}}\cdot\sec^{2}(5\theta)\cdot 5=\frac{10\sec^{2}(5\theta)}{3\sqrt[3]{\tan(5\theta)}}$ . At $\theta=\frac{\pi}{20}$ , $5\theta=\frac{\pi}{4}$ , $\tan(5\theta)=1$ and $\sec^{2}(5\theta)=2$ : $k'\left(\frac{\pi}{20}\right)=\frac{20}{3}\approx 6.667$ . The negative exponent $-\frac{1}{3}$ says what the radical was hiding: at $\theta=0$ , $\tan(5\theta)=0$ and the derivative does not exist. The function equals 0 there and is continuous, but its graph has a CUSP there, the same one as $\sqrt[3]{x^{2}}$ at the origin.

g)

The two terms look alike and have nothing in common. In $\cos(x^{2})$ , the square is INSIDE: the outer function is the cosine, hence $-\sin(x^{2})\cdot 2x$ . In $\cos^{2}x=(\cos x)^{2}$ , the square is OUTSIDE: the outer function is the power, hence $2\cos x\cdot(-\sin x)=-\sin(2x)$ . So $s'(x)=-2x\sin(x^{2})-\sin(2x)$ and $s'(1)=-2\sin 1-\sin 2\approx-1.683-0.909=-2.592$ . The reflex: before differentiating, rewrite $\cos^{2}x$ as $(\cos x)^{2}$ ; the parentheses say which function is outside.

Exercise 2: The exponential: what the base changes, and what it does not

All exponentials look alike: they pass through the point (0, 1) and never take the value zero. Only one is its own derivative, and that is what defines the number $e$ .

a) Differentiate $f(x)=e^{3x}$ , $g(x)=e^{-x^{2}}$ and $h(x)=xe^{-x}$ . For $h$ , find the point where the tangent line is horizontal.
b) The general formula is $(a^{x})'=a^{x}\ln a$ . Differentiate $u(x)=2^{x}$ and $v(x)=3^{x^{2}}$ . What does the formula become for $a=e$ ?
c) The figure shows the three curves $y=2^{x}$ , $y=e^{x}$ and $y=3^{x}$ with their tangent lines at 0. Give the slope of each and say in one sentence what characterizes the number $e$ .
d) Many students write $(e^{x})^{2}=e^{x^{2}}$ . Differentiate both expressions and show, by a computation at $x=2$ , that they have nothing in common.

Show the solution

Answers

a) $3e^{3x}$ ; $-2xe^{-x^{2}}$ ; $e^{-x}(1-x)$ , horizontal tangent at $(1, e^{-1})$
b) $2^{x}\ln 2$ ; $2x\ln 3\cdot 3^{x^{2}}$ ; for $a=e$ , $(e^{x})'=e^{x}$
c) $\ln 2$ , $1$ , $\ln 3$ : $e$ is the base with slope 1 at 0
d) $2e^{2x}$ against $2xe^{x^{2}}$ : $109.2$ against $218.4$ at $2$

a)

$f'(x)=3e^{3x}$ : chain rule, the derivative of the inside $3x$ comes out as a factor. $g'(x)=e^{-x^{2}}\cdot(-2x)=-2xe^{-x^{2}}$ : this is the Gaussian, and its derivative is zero at $x=0$ , which is its maximum. $h'(x)=1\cdot e^{-x}+x\cdot(-e^{-x})=e^{-x}(1-x)$ : product rule, then factor out the exponential, always. Since $e^{-x}>0$ , the derivative is zero only at $x=1$ , where $h(1)=e^{-1}\approx 0.368$ . It is a maximum: $h'>0$ before, $h'<0$ after.

b)

$u'(x)=2^{x}\ln 2\approx 0.693\cdot 2^{x}$ : the factor $\ln 2$ is the one most often forgotten, and it costs the whole slope computation. $v(x)=3^{x^{2}}$ needs the chain rule ON TOP of the formula: $v'(x)=3^{x^{2}}\ln 3\cdot 2x=2x\ln 3\cdot 3^{x^{2}}$ . For $a=e$ , $\ln e=1$ and the formula gives back $(e^{x})'=e^{x}$ : the general formula contains the special case, it is not one more rule to memorize.

c)

The slopes at 0 are $\ln 2\approx 0.693$ , $\ln e=1$ and $\ln 3\approx 1.099$ . This is exactly the characterization of $e$ : it is the ONLY base whose exponential curve crosses the point (0, 1) with slope 1, hence the only exponential equal to its own derivative. The other bases are horizontally stretched or compressed versions of that one, since $a^{x}=e^{x\ln a}$ .

d)

$(e^{x})^{2}=e^{2x}$ by the laws of exponents, so its derivative is $2e^{2x}$ . On the other hand, $e^{x^{2}}$ has derivative $2xe^{x^{2}}$ . At $x=2$ : $2e^{4}\approx 109.2$ against $4e^{4}\approx 218.4$ ; and the functions themselves are $e^{4}\approx 54.6$ against $e^{4}\approx 54.6$ , equal at $x=0$ and at $x=2$ only, the two solutions of $2x=x^{2}$ . At $x=3$ they equal $e^{6}\approx 403$ and $e^{9}\approx 8103$ : a factor of 20. Remember the reading: the exponent of a power of a power gets MULTIPLIED, not composed.

Exercise 3: The logarithm, and the domain that comes before the derivative

The formula $(\ln u)'=\frac{u'}{u}$ fits in five characters, and yet this is the chapter where the most marks are lost: because people differentiate before looking at where the function exists.

a) Differentiate $f(x)=\ln(3x)$ , $g(x)=\ln(x^{2}+1)$ and $h(x)=\ln(\cos x)$ , giving the domain each time.
b) You should find the same derivative for $\ln(3x)$ and for $\ln x$ . Explain why this is not a mistake.
c) Differentiate $u(x)=\log_{2}x$ and $v(x)=x\ln x-x$ . What do you notice about $v'$ ?
d) Prove that $(\ln|x|)'=\frac{1}{x}$ for every $x\neq 0$ , treating the case $x<0$ separately.
e) Give the domain of $w(x)=\ln(x^{2}-4)$ , then its derivative. Why does the expression obtained only make sense on that domain, even though it can be computed formally elsewhere?
f) Differentiate $F(x)=\ln(\sec x+\tan x)$ , $G(x)=\ln(\csc x-\cot x)$ and $L(x)=\ln(\sin x)$ on $\left(0,\frac{\pi}{2}\right)$ , simplifying all the way. What do you notice?

Type your answers, the page tells you right or wrong 0/12

a)

f'(x)

h'(x)

b)

\ln(3x)-\ln x

c)

u'(1)

v'(x)

d)

(\ln|x|)'

at

-2

e)

Left interval ,

Right interval ,

Does

w

have an extremum at

0

?

f)

F'(x)

G'(x)

L'(x)

Show the solution

Answers

a) $\frac{1}{x}$ for $x>0$ ; $\frac{2x}{x^{2}+1}$ on $\mathbb{R}$ ; $-\tan x$ where $\cos x>0$
b) $\ln(3x)=\ln 3+\ln x$ : same derivative
c) $\frac{1}{x\ln 2}$ ; $v'(x)=\ln x$
d) $(\ln(-x))'=\frac{1}{x}$ for $x<0$
e) $(-\infty,-2)\cup(2,+\infty)$ ; $w'(x)=\frac{2x}{x^{2}-4}$
f) $F'(x)=\sec x$ ; $G'(x)=\csc x$ ; $L'(x)=\cot x$

a)

$f'(x)=\frac{3}{3x}=\frac{1}{x}$ , for $x>0$ . $g'(x)=\frac{2x}{x^{2}+1}$ , defined on all of $\mathbb{R}$ since $x^{2}+1>0$ always: this is the rare logarithm with no restriction. $h'(x)=\frac{-\sin x}{\cos x}=-\tan x$ , on the intervals where $\cos x>0$ , for example $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ . The logarithm requires a STRICTLY positive argument, not merely a nonzero one.

b)

Because $\ln(3x)=\ln 3+\ln x$ : the two functions differ by a CONSTANT, and two functions that differ by a constant have the same derivative. Geometrically, the graph of $\ln(3x)$ is that of $\ln x$ shifted vertically by $\ln 3\approx 1.099$ : a vertical shift changes no slope. This is the first useful example of the theorem “zero derivative, so constant function”, which will come back in the chapter on antiderivatives.

c)

$u(x)=\log_{2}x=\frac{\ln x}{\ln 2}$ , so $u'(x)=\frac{1}{x\ln 2}\approx\frac{1.4427}{x}$ : changing the base only multiplies by a constant. $v'(x)=\left(1\cdot\ln x+x\cdot\frac{1}{x}\right)-1=\ln x+1-1=\ln x$ . In other words, $v$ is an antiderivative of $\ln$ , a result that will be proved again by integration by parts in 201-SN3-RE, integral calculus: it is useful to know right now that it can be checked in one line by differentiation.

d)

For $x>0$ , $|x|=x$ and $(\ln x)'=\frac{1}{x}$ . For $x<0$ , $|x|=-x$ , so $\ln|x|=\ln(-x)$ and by the chain rule $(\ln(-x))'=\frac{-1}{-x}=\frac{1}{x}$ : the minus sign of the inside and that of the denominator cancel. Both pieces give the same formula, hence the result for every $x\neq 0$ . Check at $x=-2$ : $\frac{\ln|-2+0.000001|-\ln|-2|}{0.000001}\approx-0.5=\frac{1}{-2}$ ✓. This is why the antiderivative of $\frac{1}{x}$ is written $\ln|x|$ and not $\ln x$ .

e)

$x^{2}-4>0$ is equivalent to $|x|>2$ , so the domain is $(-\infty,-2)\cup(2,+\infty)$ . The derivative is $w'(x)=\frac{2x}{x^{2}-4}$ . This fraction CAN be computed at $x=0$ and gives 0, but $w(0)=\ln(-4)$ does not exist: an algebraic expression knows nothing about the domain of the function it came from. The derivative of a function is never defined outside the domain of the function, even when the formula itself accepts the number. This is the mistake that leads to an extremum at $x=0$ on an interval where the function does not exist.

f)

$(\ln u)'=\frac{u'}{u}$ each time, then look for the common factor. $F'(x)=\frac{\sec x\tan x+\sec^{2}x}{\sec x+\tan x}=\frac{\sec x\left(\tan x+\sec x\right)}{\sec x+\tan x}=\sec x$ . $G'(x)=\frac{-\csc x\cot x+\csc^{2}x}{\csc x-\cot x}=\frac{\csc x\left(\csc x-\cot x\right)}{\csc x-\cot x}=\csc x$ : the cotangent and the cosecant each carry a minus sign, and they offset each other in the numerator. $L'(x)=\frac{\cos x}{\sin x}=\cot x$ , the twin of the $-\tan x$ of question a. What to notice: the derivative of the inside is a multiple of the inside, so the fraction SIMPLIFIES, and an answer left as a quotient loses the simplification mark. Read backwards, these three lines give the antiderivatives of $\sec x$ , of $\csc x$ and of $\cot x$ , which will come back in 201-SN3-RE. Domain: on $\left(0,\frac{\pi}{2}\right)$ , $\sec x+\tan x>0$ , $\sin x>0$ and $\csc x-\cot x=\frac{1-\cos x}{\sin x}>0$ , so all three logarithms exist.

Exercise 4: Nesting three layers without losing a factor

The chain rule never gets tired: you can apply it as many times as there are nested functions. The only discipline is to name the layers from the outside in BEFORE writing anything.

a) Differentiate $f(x)=e^{\sin(2x)}$ , naming the three layers.
b) Differentiate $g(x)=\ln(\tan x)$ and simplify the result as far as possible.
c) Differentiate $h(x)=\sqrt{1+e^{2x}}$ .
d) Differentiate $k(x)=\sin\left(\ln(x^{2}+1)\right)$ .
e) Differentiate $p(x)=(\ln x)^{3}$ and $q(x)=\ln(x^{3})$ . Compare the two results at $x=e$ and say what the notation hides.

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Answers

a) $2\cos(2x)e^{\sin(2x)}$
b) $\frac{2}{\sin(2x)}$
c) $\frac{e^{2x}}{\sqrt{1+e^{2x}}}$
d) $\cos\left(\ln(x^{2}+1)\right)\cdot\frac{2x}{x^{2}+1}$
e) $\frac{3(\ln x)^{2}}{x}$ and $\frac{3}{x}$ : equal at $e$ , not at $e^{2}$

a)

Three layers: the exponential, then the sine, then $2x$ . $f'(x)=e^{\sin(2x)}\cdot\cos(2x)\cdot 2=2\cos(2x)e^{\sin(2x)}$ . The factor 2 from the innermost layer is the one that gets forgotten: check at $x=0$ , $f'(0)=2\cdot 1\cdot e^{0}=2$ , whereas without the factor you would find 1, a slope half as large as it should be.

b)

$g'(x)=\frac{(\tan x)'}{\tan x}=\frac{\sec^{2}x}{\tan x}=\frac{1}{\cos^{2}x}\cdot\frac{\cos x}{\sin x}=\frac{1}{\sin x\cos x}$ . You can go further: $\sin x\cos x=\frac{1}{2}\sin(2x)$ , so $g'(x)=\frac{2}{\sin(2x)}$ . Check at $x=0.7$ : $\frac{2}{\sin 1.4}\approx\frac{2}{0.9854}\approx 2.0296$ , and the numerical rate of change gives the same value. Simplifying is not cosmetic: this is the form that will make the integral recognizable later.

c)

$h(x)=(1+e^{2x})^{1/2}$ , so $h'(x)=\frac{1}{2}(1+e^{2x})^{-1/2}\cdot 2e^{2x}=\frac{e^{2x}}{\sqrt{1+e^{2x}}}$ . The two factors of 2 cancel, which is exactly why you must write both: making them vanish in your head is the best way to lose one.

d)

$k'(x)=\cos\left(\ln(x^{2}+1)\right)\cdot\frac{2x}{x^{2}+1}$ . Three layers again, and no simplification possible: the answer stays a product, which is normal and does not signal a mistake.

e)

$p(x)=(\ln x)^{3}$ is a POWER whose inside is a logarithm: $p'(x)=3(\ln x)^{2}\cdot\frac{1}{x}=\frac{3(\ln x)^{2}}{x}$ . On the other hand, $q(x)=\ln(x^{3})=3\ln x$ is a logarithm whose inside is a power, and the property of the logarithm reduces it to a constant times $\ln x$ : $q'(x)=\frac{3}{x}$ . At $x=e$ : $p'(e)=\frac{3}{e}\approx 1.104$ and $q'(e)=\frac{3}{e}\approx 1.104$ , equal by accident since $(\ln e)^{2}=1$ . At $x=e^{2}$ : $p'=\frac{3\cdot 4}{e^{2}}\approx 1.624$ against $q'=\frac{3}{e^{2}}\approx 0.406$ , a factor of 4. The notation $\ln^{3}x$ , which you sometimes meet, means $p$ and not $q$ : exactly the kind of ambiguity that justifies writing the parentheses.

Exercise 5: Logarithmic differentiation: when no rule applies

There is one family that neither the power rule nor the exponential rule can handle: the one where the base AND the exponent contain the variable. The workaround is always the same, take the logarithm of both sides before differentiating.

a) Why do neither $(x^{n})'=nx^{n-1}$ nor $(a^{x})'=a^{x}\ln a$ apply to $y=x^{x}$ ? Compute $y'$ by logarithmic differentiation.
b) Same method for $y=x^{\sin x}$ , for $x>0$ .
c) Differentiate $y=\frac{(x^{2}+1)^{3}\sqrt{x-1}}{(2x+5)^{4}}$ by logarithmic differentiation, and say in one sentence why the direct method would be painful.
d) Give an example where logarithmic differentiation is possible but pointless, and explain the criterion for choosing.

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Answers

a) $y'=x^{x}(\ln x+1)$
b) $y'=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right)$
c) $y'=y\left(\frac{6x}{x^{2}+1}+\frac{1}{2(x-1)}-\frac{8}{2x+5}\right)$
d) $x^{5}e^{x}$ : the product rule is enough

a)

The power rule requires a CONSTANT exponent; the exponential rule requires a CONSTANT base. Here both vary, so this is a new case. Set $y=x^{x}$ with $x>0$ and take the logarithm: $\ln y=x\ln x$ . Differentiate both sides with respect to $x$ , the left side by the chain rule since $y$ depends on $x$ : $\frac{y'}{y}=\ln x+x\cdot\frac{1}{x}=\ln x+1$ . So $y'=x^{x}(\ln x+1)$ . Check at $x=2$ : $y'=4(\ln 2+1)\approx 4\cdot 1.693=6.773$ , and the numerical rate $\frac{2.000001^{2.000001}-2^{2}}{0.000001}\approx 6.773$ ✓. Along the way, $y'=0$ gives $\ln x=-1$ , that is $x=e^{-1}\approx 0.368$ : this is the minimum of $x^{x}$ , where the function equals about $0.6922$ .

b)

$\ln y=\sin x\cdot\ln x$ , then $\frac{y'}{y}=\cos x\ln x+\frac{\sin x}{x}$ , so $y'=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right)$ . The structure is always the same: the result is the function itself multiplied by the derivative of its logarithm. This is what is called the LOGARITHMIC derivative, $\frac{y'}{y}$ , and it measures a RELATIVE rate of change, in percent per unit, not in units per unit.

c)

$\ln y=3\ln(x^{2}+1)+\frac{1}{2}\ln(x-1)-4\ln(2x+5)$ : the logarithm turns the product into a sum, the quotient into a difference and the exponents into coefficients, which replaces a quotient rule nested inside a product rule by three independent derivatives. Differentiate: $\frac{y'}{y}=\frac{6x}{x^{2}+1}+\frac{1}{2(x-1)}-\frac{8}{2x+5}$ , then multiply by $y$ . Check at $x=2$ : $y=\frac{125\cdot 1}{9^{4}}=\frac{125}{6561}\approx 0.019052$ and $\frac{y'}{y}=\frac{12}{5}+\frac{1}{2}-\frac{8}{9}\approx 2.0111$ , so $y'\approx 0.038318$ . The numerical rate confirms it. Directly, you would have to differentiate a quotient whose numerator is itself a product of two composite functions: three pages, and a sign error almost guaranteed.

d)

For $y=x^{5}e^{x}$ , logarithmic differentiation works and gives $\frac{y'}{y}=\frac{5}{x}+1$ , so $y'=x^{5}e^{x}\left(\frac{5}{x}+1\right)=x^{4}e^{x}(5+x)$ ; but the product rule gives the same result in one line. The criterion: switch to the logarithm when the function is a PRODUCT OR QUOTIENT OF AT LEAST THREE FACTORS with exponents, or when the variable is in the exponent. Two simple factors do not justify it, and the detour then costs more than the direct rule.

Exercise 6: The inverse trigonometric functions and their algebraic derivatives

The remarkable fact of this section: the inverses of trigonometric functions have ALGEBRAIC derivatives, with no trigonometry at all. This is why they will show up as antiderivatives of fractions in 201-SN3-RE, integral calculus.

a) Let $y=\arcsin x$ , so $\sin y=x$ with $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ . By differentiating both sides with respect to $x$ , prove that $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ , justifying the choice of sign of the square root.
b) By the same method, prove that $(\arctan x)'=\frac{1}{1+x^{2}}$ .
c) Differentiate $f(x)=\arctan(2x)$ , $g(x)=\arcsin\left(\frac{x}{3}\right)$ and $h(x)=x\arctan x-\frac{1}{2}\ln(1+x^{2})$ . Comment on the result obtained for $h$ .
d) The figure shows $y=\arctan x$ . Read from it how the derivative behaves as $x\to\pm\infty$ , then explain why $(\arcsin x)'$ blows up at $x=\pm 1$ even though $\arcsin$ is perfectly well defined there.
e) Give the domain of $p(x)=\arccos\left(\frac{3}{x}\right)$ , compute $p'(x)$ while simplifying the square root, then $p'(-5)$ . Same work for $q(x)=\arcsin(\sqrt{x})$ : domain, derivative, then $q'\left(\frac{1}{4}\right)$ .

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

a)

The sign of the square root comes from

b)

(\arctan x)'

c)

f'(x)

g'(x)

h'(x)

d)

Maximum slope of

\arctan

\arcsin

at

x=1

e)

Domain of

p

p'(-5)

Domain of

q

,

q'\left(\frac{1}{4}\right)

Show the solution

Answers

a) $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ , positive root because $\cos y\ge 0$
b) $(\arctan x)'=\frac{1}{1+x^{2}}$
c) $\frac{2}{1+4x^{2}}$ ; $\frac{1}{\sqrt{9-x^{2}}}$ ; $h'(x)=\arctan x$
d) Slope tends to $0$ at infinity; vertical tangent line of $\arcsin$ at $\pm 1$
e) Domain $|x|\ge 3$ ; $p'(x)=\frac{3}{|x|\sqrt{x^{2}-9}}$ , $p'(-5)=0.15$ ; $q'(x)=\frac{1}{2\sqrt{x(1-x)}}$ on $(0,1)$ , $q'\left(\frac{1}{4}\right)\approx 1.155$

a)

Differentiate $\sin y=x$ with respect to $x$ , treating $y$ as a function of $x$ : $\cos y\cdot y'=1$ , so $y'=\frac{1}{\cos y}$ . It remains to express $\cos y$ in terms of $x$ : from $\sin^{2}y+\cos^{2}y=1$ we get $\cos y=\pm\sqrt{1-x^{2}}$ . The sign is decided by the chosen DOMAIN: on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , the cosine is positive or zero, so we keep the positive root and $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ for $|x|<1$ . The sign never comes from the computation; it comes from the interval on which the function was made invertible.

b)

From $\tan y=x$ we get $\sec^{2}y\cdot y'=1$ , so $y'=\cos^{2}y$ . Now $\sec^{2}y=1+\tan^{2}y=1+x^{2}$ , hence $y'=\frac{1}{1+x^{2}}$ . This time there is no choice of sign: the result is the reciprocal of a square, positive everywhere, which confirms that $\arctan$ is strictly increasing on all of $\mathbb{R}$ .

c)

$f'(x)=\frac{2}{1+4x^{2}}$ by the chain rule. $g'(x)=\frac{1/3}{\sqrt{1-\frac{x^{2}}{9}}}=\frac{1}{\sqrt{9-x^{2}}}$ after bringing the 3 under the root, a much more usable form. For $h$ : $h'(x)=\arctan x+\frac{x}{1+x^{2}}-\frac{1}{2}\cdot\frac{2x}{1+x^{2}}=\arctan x$ . The last two terms cancel exactly: $h$ is therefore an antiderivative of $\arctan$ , which will be proved again by integration by parts in the next course. Check at $x=1$ : $h'(1)=\arctan 1=\frac{\pi}{4}\approx 0.7854$ , and the numerical rate of change gives the same value.

d)

In the figure, the curve flattens out towards its two asymptotes $y=\pm\frac{\pi}{2}$ : the slope tends to 0, as confirmed by $\frac{1}{1+x^{2}}\to 0$ . The maximum slope is 1 and it is reached at $x=0$ . For $\arcsin$ , as $x\to 1^{-}$ the denominator $\sqrt{1-x^{2}}\to 0^{+}$ , so the derivative tends to $+\infty$ : the graph of $\arcsin$ has a VERTICAL TANGENT LINE at $x=1$ . There is no contradiction; a function can be continuous and defined at a point without being differentiable there. This is the mirror reading of the graph of $\sin$ , whose tangent line is horizontal at $\frac{\pi}{2}$ : swapping the axes swaps horizontal and vertical.

e)

$\arccos$ requires an argument in $[-1,1]$ : $\left|\frac{3}{x}\right|\le 1$ is equivalent to $|x|\ge 3$ , so the domain is $\left(-\infty,-3\right]\cup\left[3,+\infty\right)$ , and the derivative exists for $|x|>3$ . Chain rule: $p'(x)=\frac{-1}{\sqrt{1-\frac{9}{x^{2}}}}\cdot\left(-\frac{3}{x^{2}}\right)=\frac{3}{x^{2}\sqrt{1-\frac{9}{x^{2}}}}$ . To simplify, $\sqrt{1-\frac{9}{x^{2}}}=\frac{\sqrt{x^{2}-9}}{\sqrt{x^{2}}}=\frac{\sqrt{x^{2}-9}}{|x|}$ , and this is where everything is decided: $\sqrt{x^{2}}=|x|$ , not $x$ . Hence $p'(x)=\frac{3|x|}{x^{2}\sqrt{x^{2}-9}}=\frac{3}{|x|\sqrt{x^{2}-9}}$ , and $p'(-5)=\frac{3}{5\cdot 4}=0.15$ . The wrong form $\frac{3}{x\sqrt{x^{2}-9}}$ would give $-0.15$ , a negative slope for an INCREASING function: for $x<-3$ , $\frac{3}{x}$ decreases and $\arccos$ decreases, so $p$ increases. For $q$ , you need $x\ge 0$ for the root and $\sqrt{x}\le 1$ for $\arcsin$ : the domain is $[0,1]$ . Then $q'(x)=\frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x(1-x)}}$ , defined on $(0,1)$ only: vertical tangent lines at both endpoints. Finally $q'\left(\frac{1}{4}\right)=\frac{1}{2\sqrt{\frac{3}{16}}}=\frac{2}{\sqrt{3}}\approx 1.155$ .

Exercise 7: The hyperbolic functions and the shape of the catenary

The hyperbolic cosine is not a parabola, even though it looks like one: it is the shape taken by a cable hanging under its own weight. Its derivatives are those of trigonometry, but without the minus sign.

a) Let $\cosh x=\frac{e^{x}+e^{-x}}{2}$ and $\sinh x=\frac{e^{x}-e^{-x}}{2}$ . Compute their derivatives and compare with the pair $(\cos,\sin)$ .
b) Prove the identity $\cosh^{2}x-\sinh^{2}x=1$ , then differentiate the left side to check that it is consistent with question a.
c) Let $\tanh x=\frac{\sinh x}{\cosh x}$ . Show that $(\tanh x)'=\frac{1}{\cosh^{2}x}=1-\tanh^{2}x$ , and deduce whether $\tanh$ is increasing or decreasing.
d) A cable hanging between two poles takes the shape $y=a\cosh\left(\frac{x}{a}\right)$ with $a=20$ metres, the origin being at the lowest point of the cable. Give the slope of the cable at $x=20$ metres, and how high it has risen above its lowest point.
e) The figure overlays $y=\cosh x$ and the parabola $y=1+\frac{x^{2}}{2}$ . What do they share at $x=0$ , and where do they part ways?

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Answers

a) $(\cosh x)'=\sinh x$ , $(\sinh x)'=\cosh x$
b) $\cosh^{2}x-\sinh^{2}x=1$
c) $(\tanh x)'=\frac{1}{\cosh^{2}x}>0$ : increasing
d) Slope $\sinh 1\approx 1.175$ ; $10.86$ m
e) Same value, slope and concavity at $0$ ; gap of $0.762$ at $2$

a)

$(\cosh x)'=\frac{e^{x}-e^{-x}}{2}=\sinh x$ and $(\sinh x)'=\frac{e^{x}+e^{-x}}{2}=\cosh x$ : each one is the derivative of the other, WITHOUT a minus sign. That is the whole difference from the trigonometric pair, where $(\cos x)'=-\sin x$ . The consequence is structural: $\cos$ and $\sin$ satisfy $y''=-y$ and oscillate, while $\cosh$ and $\sinh$ satisfy $y''=+y$ and blow up. A single sign separates a wave from exponential growth.

b)

$\cosh^{2}x-\sinh^{2}x=\frac{(e^{x}+e^{-x})^{2}-(e^{x}-e^{-x})^{2}}{4}=\frac{4e^{x}e^{-x}}{4}=1$ , using the identity $A^{2}-B^{2}=(A-B)(A+B)$ or simply expanding. Check by differentiation: $\left(\cosh^{2}x-\sinh^{2}x\right)'=2\cosh x\sinh x-2\sinh x\cosh x=0$ , and a function with zero derivative on $\mathbb{R}$ is constant; its value at 0 is $1-0=1$ . The two routes agree, which validates the derivatives of question a.

c)

By the quotient rule: $(\tanh x)'=\frac{\cosh x\cosh x-\sinh x\sinh x}{\cosh^{2}x}=\frac{\cosh^{2}x-\sinh^{2}x}{\cosh^{2}x}=\frac{1}{\cosh^{2}x}$ . Splitting the fraction term by term instead, you get $1-\tanh^{2}x$ : this is the exact copy of what was done for $\tan$ in exercise 1. Since $\cosh x\geq 1>0$ , the derivative is strictly positive: $\tanh$ is strictly increasing on $\mathbb{R}$ , and its maximum slope is 1, at $x=0$ . It is bounded between $-1$ and 1, which makes it the classic activation function of neural networks.

d)

$y'=a\sinh\left(\frac{x}{a}\right)\cdot\frac{1}{a}=\sinh\left(\frac{x}{a}\right)$ : the parameter $a$ disappears from the slope, a neat result. At $x=20$ with $a=20$ : $y'=\sinh(1)=\frac{e-e^{-1}}{2}\approx 1.1752$ , an inclination of about $\arctan(1.1752)\approx 49.6$ degrees. The height is $y(20)-y(0)=20\cosh(1)-20\cosh(0)=20(1.5431-1)\approx 10.86$ metres. So the cable has risen a little less than 11 metres at 20 metres from the lowest point.

e)

At $x=0$ the two curves both equal 1, have zero slope and the SAME concavity: $(\cosh)''(0)=\cosh 0=1$ and $\left(1+\frac{x^{2}}{2}\right)''=1$ . They therefore agree up to order 2, which explains why the catenary was long believed to be a parabola. They part ways at order 4: $\cosh x=1+\frac{x^{2}}{2}+\frac{x^{4}}{24}+\dots$ , so the gap is about $\frac{x^{4}}{24}$ , negligible near 0 but visible from $x=2$ on, where $\cosh 2\approx 3.762$ against $3$ for the parabola.

Exercise 8: Five statements to correct

Each of the following five statements is FALSE. Say why and give the correct statement.

1) “Since $(x^{n})'=nx^{n-1}$ , we have $(2^{x})'=x\,2^{x-1}$ .”
2) “ $\left(\ln(x^{2})\right)'=\frac{1}{x^{2}}$ , since the derivative of the logarithm is the reciprocal of its argument.”
3) “ $\left(\sin^{2}x\right)'=\cos^{2}x$ , by analogy with the derivative of sine.”
4) “ $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ is defined for every real $x$ , since it is a fraction.”
5) “ $\cosh$ and $\cos$ differ only by a sign, so $(\cosh x)'=-\sinh x$ .”

Show the solution

Answers

1) $(2^{x})'=2^{x}\ln 2$
2) $\left(\ln(x^{2})\right)'=\frac{2}{x}$
3) $\left(\sin^{2}x\right)'=\sin(2x)$
4) Defined on $(-1,1)$ only
5) $(\cosh x)'=\sinh x$

1)

FALSE. The power rule requires a CONSTANT exponent and a variable base; here it is the other way around, the base 2 is constant and the exponent varies. The correct formula is $(a^{x})'=a^{x}\ln a$ , so $(2^{x})'=2^{x}\ln 2$ . The numerical check settles it: at $x=3$ , the false formula gives $3\cdot 2^{2}=12$ , the true one gives $8\ln 2\approx 5.545$ , and the numerical rate of change is indeed about $5.545$ .

2)

FALSE. The derivative of the logarithm is not “the reciprocal of the argument”, it is $\frac{u'}{u}$ : the derivative of the inside is missing. Here $u=x^{2}$ so $u'=2x$ and $\left(\ln(x^{2})\right)'=\frac{2x}{x^{2}}=\frac{2}{x}$ . You can get it without computation by simplifying first: $\ln(x^{2})=2\ln|x|$ , with derivative $\frac{2}{x}$ . At $x=2$ : $\frac{2}{2}=1$ , whereas the statement claims $0.25$ .

3)

FALSE. $\sin^{2}x$ means $(\sin x)^{2}$ , a POWER, so it is the chain rule that applies: $\left(\sin^{2}x\right)'=2\sin x\cos x=\sin(2x)$ . The analogy invoked has no basis; you do not differentiate a square by differentiating what is inside it. At $x=1$ : $\sin 2\approx 0.909$ , whereas $\cos^{2}1\approx 0.292$ .

4)

FALSE. A fraction is defined only if its denominator is nonzero AND if the square root exists: you need $1-x^{2}>0$ , so $-1<x<1$ . The derivative is defined only on the INTERIOR of the domain of $\arcsin$ , which is $[-1,1]$ : at both endpoints, $\arcsin$ exists but has a vertical tangent line. Confusing the domain of a function with that of its derivative is the most costly mistake of the chapter.

5)

FALSE. The minus sign appears with $\cos$ , not with $\cosh$ : $(\cos x)'=-\sin x$ but $(\cosh x)'=+\sinh x$ . You can check it directly on the definition, $\left(\frac{e^{x}+e^{-x}}{2}\right)'=\frac{e^{x}-e^{-x}}{2}=\sinh x$ . This sign is not a detail of notation: it is what makes $\cos$ oscillate and $\cosh$ grow without bound.

Exercise 9: Problem: radioactive decay and the relative rate

A quantity that decreases in proportion to itself follows an exponential. All of carbon-14 dating fits in that sentence and in one derivative.

a) The mass of a carbon-14 sample follows $N(t)=N_{0}e^{-\lambda t}$ , where $t$ is in years. Compute $N'(t)$ and show that the ratio $\frac{N'(t)}{N(t)}$ is CONSTANT. What does this constant mean physically?
b) The half-life of carbon 14 is 5730 years. Find $\lambda$ , then the percentage lost per year.
c) A bone fragment contains 22 percent of the carbon 14 of a living bone of the same mass. Date it.
d) A coffee at 85 degrees Celsius in a room at 20 degrees cools according to Newton's law $T(t)=20+65e^{-kt}$ , with $t$ in minutes. It is at 60 degrees after 10 minutes. Compute $k$ , then the cooling rate at that moment, in degrees per minute.
e) Why is the cooling rate of the coffee NOT constant even though the relative rate of question a) is? Compare the two situations.

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Answers

a) $\frac{N'}{N}=-\lambda$
b) $\lambda\approx 1.2097\times 10^{-4}$ per year; $0.0121$ % per year
c) About $12\,500$ years
d) $k\approx 0.0486$ per minute; $T'(10)\approx-1.94$ °C/min
e) It is $\frac{T'}{T-20}$ that is constant

a)

$N'(t)=N_{0}e^{-\lambda t}\cdot(-\lambda)=-\lambda N(t)$ , so $\frac{N'(t)}{N(t)}=-\lambda$ , constant and independent of $t$ as well as of $N_{0}$ . Physically: every year, the same PROPORTION of nuclei decays, whatever the quantity left. The quantity lost, on the other hand, decreases with time, which is exactly why exponential decay never reaches zero. This is the very definition of the exponential model: the derivative is proportional to the function.

b)

The half-life $T$ satisfies $e^{-\lambda T}=\frac{1}{2}$ , so $\lambda=\frac{\ln 2}{T}=\frac{0.6931}{5730}\approx 1.2097\times 10^{-4}$ per year. The relative loss per year is therefore about $0.0121$ percent, roughly one nucleus in 8,267 per year. Notice that the computation does not depend on any initial quantity, which is what makes dating possible without knowing the original mass.

c)

Solve $e^{-\lambda t}=0.22$ , that is $t=\frac{-\ln(0.22)}{\lambda}=\frac{1.5141}{1.2097\times 10^{-4}}\approx 12\,516$ years. Round honestly to about 12,500 years: the precision of the 22 percent measurement does not justify four significant figures. An error of one percentage point in the measurement moves the date by more than 350 years, an order of magnitude that a serious solution should point out.

d)

$T(10)=60$ gives $20+65e^{-10k}=60$ , that is $e^{-10k}=\frac{40}{65}=0.61538$ , hence $k=\frac{-\ln(0.61538)}{10}\approx 0.048551$ per minute. The rate is $T'(t)=-65k\,e^{-kt}=-k\,(T(t)-20)$ ; at $t=10$ , $T'(10)=-0.048551\times 40\approx-1.942$ degrees per minute. The negative sign says that the temperature is dropping, and writing $T'=-k(T-20)$ shows that it is the GAP from the room temperature, not the temperature itself, that drives the cooling.

e)

In both cases the derivative is proportional to something, but not to the same quantity. For carbon, $N'=-\lambda N$ : the reference is zero, so the relative rate $\frac{N'}{N}$ is constant. For the coffee, $T'=-k(T-20)$ : the reference is 20 degrees, not zero, so it is $\frac{T'}{T-20}$ that is constant, while $\frac{T'}{T}$ varies. At the start the coffee loses $-0.048551\times 65\approx-3.16$ degrees per minute, against $-1.94$ after ten minutes: the cooling rate has dropped by nearly 40 percent while the rate relative to the gap has not moved. The lesson goes beyond these two models: before saying “the rate is constant”, say relative to WHAT.

Exercise 10: Problem: the logistic model and its maximum rate of spread

An exponential grows without bound, which no real population does. The logistic model fixes this, and the derivative answers the real question: when is the spread fastest?

a) A rumour spreads on a campus of 1000 people according to $P(t)=\frac{1000}{1+99e^{-0.6t}}$ , where $t$ is in days. How many people know it at $t=0$ , and what value does $P$ approach?
b) Compute $P'(t)$ by the chain rule applied to the power $-1$ , and check that $P'>0$ for every $t$ .
c) Show that $P'$ can be written $P'=\frac{0.6}{1000}P(1000-P)$ , then deduce from it, WITHOUT computing $P''$ , the moment when the spread is fastest and the rate of spread reached that day.
d) Confirm with $P''$ that this moment is indeed an inflection point, and read it on the figure.
e) After how many days does 90 percent of the campus know the rumour? Compare with the pure exponential model $Q(t)=10e^{0.6t}$ , which has the same starting value and almost the same starting slope.

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Answers

a) $10$ people; limit $1000$
b) $P'(t)=\frac{59\,400\,e^{-0.6t}}{\left(1+99e^{-0.6t}\right)^{2}}>0$
c) Maximum at $P=500$ : $150$ people per day, $t\approx 7.66$ days
d) $P''$ changes sign at $P=500$ : inflection point
e) $11.32$ days against $7.50$ days

a)

$P(0)=\frac{1000}{1+99}=10$ people. As $t\to+\infty$ , $e^{-0.6t}\to 0$ so $P\to 1000$ : this is the plateau, the capacity of the campus, drawn dashed in the figure below. The model is built so that the population never exceeds 1000, unlike an exponential.

b)

Write $P(t)=1000\left(1+99e^{-0.6t}\right)^{-1}$ . Chain rule: $P'(t)=-1000\left(1+99e^{-0.6t}\right)^{-2}\cdot 99e^{-0.6t}\cdot(-0.6)$ , that is $P'(t)=\frac{59\,400\,e^{-0.6t}}{\left(1+99e^{-0.6t}\right)^{2}}$ . The two minus signs cancel, so $P'>0$ for every $t$ : the rumour never retreats, which is exactly what the model should say.

c)

Setting $u=99e^{-0.6t}$ , we have $P=\frac{1000}{1+u}$ and $1000-P=\frac{1000u}{1+u}$ , so $\frac{0.6}{1000}P(1000-P)=\frac{0.6}{1000}\cdot\frac{1000}{1+u}\cdot\frac{1000u}{1+u}=\frac{600u}{(1+u)^{2}}$ , which is exactly the expression of $P'$ obtained in b. This form says much more: $P'$ is a product of two factors whose SUM is constant, equal to 1000. Such a product is largest when the two factors are equal, so when $P=500$ , half the campus. The rate of spread is then $\frac{0.6}{1000}\times 500\times 500=150$ people per day. The moment is found by solving $99e^{-0.6t}=1$ , that is $t=\frac{\ln 99}{0.6}\approx 7.659$ days.

d)

Differentiating $P'=\frac{0.6}{1000}\left(1000P-P^{2}\right)$ gives $P''=\frac{0.6}{1000}\left(1000-2P\right)P'$ . Since $P'>0$ , the sign of $P''$ is that of $1000-2P$ : positive as long as $P<500$ , negative afterwards, with a zero and a CHANGE OF SIGN at $P=500$ . It is therefore indeed an inflection point, and the curve of the figure switches there from concave up to concave down: this is the point where it stops accelerating, marked by the dot where the two dashed lines meet in the figure below.

e)

$P(t)=900$ gives $1+99e^{-0.6t}=\frac{1000}{900}$ , that is $99e^{-0.6t}=\frac{1}{9}$ and $t=\frac{\ln 891}{0.6}\approx 11.32$ days. The exponential model $Q(t)=10e^{0.6t}$ reaches 900 at $t=\frac{\ln 90}{0.6}\approx 7.50$ days, and it would reach 1000 at $7.68$ days and then 10,000 at $11.5$ days, which makes no sense on a campus of 1000 people. Yet the two models are practically indistinguishable at the very beginning, with the same starting value and almost the same starting slope: at $t=1$ , $P\approx 18.0$ and $Q\approx 18.2$ . That is the lesson of the problem: two models can fit the same initial data and diverge completely, and it is the derivative, not the value, that reveals which of the two assumptions was really made.

Exercise 11: The tangent lines to the exponential and the logarithm that pass through the origin

The figure shows $y=e^{x}$ and $y=\ln x$ , with two dashed tangent lines that pass through the origin.

a) Write the equation of the tangent line to the curve $y=e^{x}$ at the point with $x$ -coordinate $a$ .
b) For which value of $a$ does this tangent line pass through the origin? Give its equation.
c) Same question for the curve $y=\ln x$ at the point with $x$ -coordinate $b$ .
d) Prove that $\ln x\le\frac{x}{e}$ for every $x>0$ , then deduce that $\pi^{e}<e^{\pi}$ .
e) At which point is the tangent line to $y=e^{2x}$ parallel to the line $y=4x+1$ ? Give its $y$ -intercept.

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Answers

a) $y=e^{a}x+e^{a}(1-a)$
b) $a=1$ : $y=ex$
c) $b=e$ : $y=\frac{x}{e}$
d) $\ln x\le\frac{x}{e}$ ; $\pi^{e}\approx 22.46<e^{\pi}\approx 23.14$
e) $x=\frac{\ln 2}{2}\approx 0.347$ ; $y=4x+2-2\ln 2$

a)

$(e^{x})'=e^{x}$ , so the slope at $a$ is $e^{a}$ and the tangent line is $y=e^{a}(x-a)+e^{a}=e^{a}x+e^{a}(1-a)$ .

b)

The tangent line passes through the origin when its $y$ -intercept is zero: $e^{a}(1-a)=0$ . An exponential is never zero, so $a=1$ . The tangent line is $y=ex$ , with slope $e\approx 2.718$ , and it touches the curve at $(1, e)$ . This is another way of meeting $e$ : the only line through the origin that just touches the exponential curve has slope $e$ .

c)

The slope at $b$ is $\frac{1}{b}$ : $y=\frac{1}{b}(x-b)+\ln b=\frac{x}{b}-1+\ln b$ . The $y$ -intercept is zero when $\ln b=1$ , that is $b=e$ . The tangent line is $y=\frac{x}{e}$ , with slope $\frac{1}{e}\approx 0.368$ , and it touches the curve at $(e, 1)$ . The two points of contact $(1, e)$ and $(e, 1)$ are symmetric about the line $y=x$ , like the two curves, which are inverses of each other, and the two slopes are reciprocals.

d)

Let $g(x)=\frac{x}{e}-\ln x$ on $(0, +\infty)$ . $g'(x)=\frac{1}{e}-\frac{1}{x}$ is negative for $x<e$ and positive for $x>e$ : $g$ decreases then increases, and its minimum is $g(e)=1-1=0$ . So $g(x)\ge 0$ , that is $\ln x\le\frac{x}{e}$ , with equality only at $x=e$ : the graph of the logarithm lies below its tangent line through the origin, the one from c). At $x=\pi$ : $\ln\pi<\frac{\pi}{e}$ , that is $e\ln\pi<\pi$ , and since the exponential is increasing, $\pi^{e}<e^{\pi}$ . Numerically, $\pi^{e}\approx 22.46$ and $e^{\pi}\approx 23.14$ .

e)

The slope of $y=e^{2x}$ is $2e^{2x}$ . Parallel to $y=4x+1$ : $2e^{2x}=4$ , that is $e^{2x}=2$ and $x=\frac{\ln 2}{2}\approx 0.347$ . The point of contact is $\left(\frac{\ln 2}{2}, 2\right)$ and the tangent line is $y=4\left(x-\frac{\ln 2}{2}\right)+2=4x+2-2\ln 2$ , with $y$ -intercept $2-2\ln 2\approx 0.614$ . Forgetting the factor $2$ from the chain rule would give $e^{2x}=4$ and a wrong point.

Exercise 12: Zero derivative, constant function: two identities of the inverse functions

Assume that a differentiable function whose derivative is zero on an INTERVAL is constant there. Recall $(\arcsin x)'=\frac{1}{\sqrt{1-x^{2}}}$ and $(\arctan x)'=\frac{1}{1+x^{2}}$ .

a) By differentiating $\cos y=x$ with $y\in[0,\pi]$ , establish $(\arccos x)'$ on $(-1,1)$ .
b) Deduce that $\arcsin x+\arccos x$ is constant on $[-1,1]$ and give its value.
c) Let $L(x)=\arcsin x-\arctan\left(\frac{x}{\sqrt{1-x^{2}}}\right)$ on $(-1, 1)$ . Compute $L'(x)$ , then $L(0)$ . What do you conclude?
d) Let $H(x)=\arctan\left(\frac{1+x}{1-x}\right)-\arctan x$ for $x\neq 1$ . Show that $H'(x)=0$ and give the value of $H$ on each of the two intervals of its domain.
e) Check the identity of c) at $x=\frac{1}{2}$ , then explain why it does not extend to $x=1$ .

Show the solution

Answers

a) $(\arccos x)'=-\frac{1}{\sqrt{1-x^{2}}}$
b) $\arcsin x+\arccos x=\frac{\pi}{2}$
c) $L'=0$ and $L(0)=0$ : $\arcsin x=\arctan\frac{x}{\sqrt{1-x^{2}}}$
d) $H=\frac{\pi}{4}$ on $(-\infty,1)$ , $-\frac{3\pi}{4}$ on $(1,+\infty)$
e) $\frac{\pi}{6}$ on both sides; meaningless at $x=1$

a)

$\cos y=x$ gives $-\sin y\cdot y'=1$ , so $y'=-\frac{1}{\sin y}$ . On $[0,\pi]$ , $\sin y\ge 0$ , so $\sin y=\sqrt{1-\cos^{2}y}=\sqrt{1-x^{2}}$ and $(\arccos x)'=-\frac{1}{\sqrt{1-x^{2}}}$ on $(-1,1)$ . The sign of the square root is chosen by the interval $[0,\pi]$ , just as that of $\arcsin$ was chosen by $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ .

b)

$(\arcsin x+\arccos x)'=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}=0$ on the interval $(-1,1)$ : the sum is constant there, and by continuity at the endpoints as well. At $x=0$ : $\arcsin 0+\arccos 0=0+\frac{\pi}{2}$ . So $\arcsin x+\arccos x=\frac{\pi}{2}$ for every $x\in[-1,1]$ : the two angles are complementary, like the two acute angles of a right triangle.

c)

Let $u=\frac{x}{\sqrt{1-x^{2}}}=x(1-x^{2})^{-1/2}$ : $u'=(1-x^{2})^{-1/2}+x^{2}(1-x^{2})^{-3/2}=(1-x^{2})^{-3/2}$ and $1+u^{2}=\frac{1}{1-x^{2}}$ . So $\left(\arctan u\right)'=\frac{u'}{1+u^{2}}=\frac{1}{\sqrt{1-x^{2}}}$ , which is also the derivative of $\arcsin$ : $L'(x)=0$ . On the interval $(-1, 1)$ , $L$ is constant, equal to $L(0)=0$ . So $\arcsin x=\arctan\frac{x}{\sqrt{1-x^{2}}}$ : this is the reading of a right triangle with hypotenuse $1$ and opposite side $x$ , whose adjacent side is $\sqrt{1-x^{2}}$ .

d)

Let $u=\frac{1+x}{1-x}$ : $u'=\frac{(1-x)+(1+x)}{(1-x)^{2}}=\frac{2}{(1-x)^{2}}$ and $1+u^{2}=\frac{(1-x)^{2}+(1+x)^{2}}{(1-x)^{2}}=\frac{2+2x^{2}}{(1-x)^{2}}$ . So $\left(\arctan u\right)'=\frac{u'}{1+u^{2}}=\frac{2}{2+2x^{2}}=\frac{1}{1+x^{2}}$ , and $H'(x)=0$ . On $(-\infty,1)$ : $H(0)=\arctan 1-0=\frac{\pi}{4}$ . On $(1,+\infty)$ : at $x=\sqrt{3}$ , $\frac{1+\sqrt{3}}{1-\sqrt{3}}=-(2+\sqrt{3})$ , whose arctangent is $-\frac{5\pi}{12}$ , and $H(\sqrt{3})=-\frac{5\pi}{12}-\frac{\pi}{3}=-\frac{3\pi}{4}$ . In other words, $\arctan\frac{1+x}{1-x}=\arctan x+\frac{\pi}{4}$ for $x<1$ , and this equality loses $\pi$ when crossing $x=1$ .

e)

$\arcsin\frac{1}{2}=\frac{\pi}{6}$ , and $\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}$ , whose arctangent is also $\frac{\pi}{6}$ ✓. At $x=1$ , $\arcsin 1=\frac{\pi}{2}$ exists, but $\frac{x}{\sqrt{1-x^{2}}}$ is no longer defined: the right side makes no sense, like the tangent of a right angle. The identity holds on the open interval where the zero derivative was established, and nowhere else.

Exercise 13: The horizontal tangent lines of e to the x times sin x

Let $f(x)=e^{x}\sin x$ , studied on $[0,2\pi]$ .

a) Compute $f'(x)$ and factor out the exponential.
b) Solve $f'(x)=0$ on $[0,2\pi]$ .
c) Give the nature and the value of each extremum, to one decimal place.
d) Show that $f(x+\pi)=-e^{\pi}f(x)$ . What does this tell you about the ratio of two consecutive extrema?
e) Compute $f''(x)$ and give the $x$ -coordinates of the inflection points on $[0,2\pi]$ .

Show the solution

Answers

a) $f'(x)=e^{x}(\sin x+\cos x)$
b) $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$
c) Maximum $\approx 7.5$ , minimum $\approx-172.6$
d) Ratio $-e^{\pi}\approx-23.14$
e) $f''(x)=2e^{x}\cos x$ ; inflection points at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$

a)

Product rule: $f'(x)=e^{x}\sin x+e^{x}\cos x=e^{x}(\sin x+\cos x)$ . Factoring out the exponential is the move that makes the equation $f'(x)=0$ readable.

b)

$e^{x}>0$ , so $f'(x)=0$ is equivalent to $\sin x+\cos x=0$ , that is $\tan x=-1$ where $\cos x\neq 0$ . On $[0,2\pi]$ : $x=\frac{3\pi}{4}$ and $x=\frac{7\pi}{4}$ . No solution is lost by dividing by $\cos x$ : if $\cos x=0$ , then $\sin x=\pm 1$ and the sum is not zero.

c)

$\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$ is positive on $\left[0,\frac{3\pi}{4}\right)$ , negative on $\left(\frac{3\pi}{4},\frac{7\pi}{4}\right)$ , positive afterwards. Maximum at $\frac{3\pi}{4}$ : $f=e^{3\pi/4}\cdot\frac{\sqrt{2}}{2}\approx 10.55\times 0.7071\approx 7.5$ . Minimum at $\frac{7\pi}{4}$ : $f=-e^{7\pi/4}\cdot\frac{\sqrt{2}}{2}\approx-244.2\times 0.7071\approx-172.6$ .

d)

$f(x+\pi)=e^{x+\pi}\sin(x+\pi)=e^{\pi}e^{x}\cdot(-\sin x)=-e^{\pi}f(x)$ . Now $f'(x+\pi)=-e^{\pi}f'(x)$ as well: the critical points repeat every $\pi$ , and each extremum is the opposite of the previous one multiplied by $e^{\pi}\approx 23.14$ . Check: $\frac{-172.6}{7.46}\approx-23.1$ . The curve oscillates like a sine, but with an amplitude that is multiplied by $23$ every half-period.

e)

$f''(x)=e^{x}(\sin x+\cos x)+e^{x}(\cos x-\sin x)=2e^{x}\cos x$ . It is zero and changes sign at $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ : two inflection points. They are NOT halfway between the extrema, unlike what happens for $\sin x$ : each one comes exactly $\frac{\pi}{4}$ before the next extremum ( $\frac{\pi}{2}$ before $\frac{3\pi}{4}$ , $\frac{3\pi}{2}$ before $\frac{7\pi}{4}$ ). Compared with $\sin x$ , the extrema move right by $\frac{\pi}{4}$ and the inflection points by $\frac{\pi}{2}$ : the growth of $e^{x}$ delays the moment when the slope changes sign.

Exercise 14: Problem: the tide and the rule of twelfths

In a harbour, the water depth, in metres, is modelled by $h(t)=3+2.5\cos\left(\frac{\pi t}{6.2}\right)$ , where $t$ is the time in hours since high tide. The figure shows a complete cycle, with the $2$ m level dashed.

a) Give the depth at high tide, at low tide, and the tidal range, the difference between the two.
b) Compute $h'(t)$ , then the rate of change of the water level one hour after high tide.
c) At what moment does the water fall fastest, and at what rate?
d) Sailors split the fall into six equal intervals and remember that it loses $\frac{1}{12}$ , $\frac{2}{12}$ , $\frac{3}{12}$ , $\frac{3}{12}$ , $\frac{2}{12}$ then $\frac{1}{12}$ of the tidal range. Compare with the model on the first and the third interval.
e) A boat needs $2$ m of water. Until what time after high tide can it leave, and how fast is the level falling at that moment?

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Answers

a) $5.5$ m, $0.5$ m, tidal range $5$ m
b) $h'(t)=-\frac{2.5\pi}{6.2}\sin\left(\frac{\pi t}{6.2}\right)$ ; $h'(1)\approx-0.61$ m/h
c) $t=3.1$ h: $1.27$ m/h
d) $0.33$ m against $0.42$ ; exactly $1.25$ m
e) $t\approx 3.91$ h; falling at $1.16$ m/h

a)

High tide: $h(0)=3+2.5=5.5$ m. Low tide when the cosine equals $-1$ , at $t=6.2$ h: $h=0.5$ m. Tidal range: $5$ m.

b)

By the chain rule, $h'(t)=-2.5\cdot\frac{\pi}{6.2}\sin\left(\frac{\pi t}{6.2}\right)\approx-1.267\sin\left(\frac{\pi t}{6.2}\right)$ . At $t=1$ : $h'(1)\approx-1.267\times 0.485\approx-0.61$ m/h. The factor $\frac{\pi}{6.2}$ , which comes from the chain rule, is the one that gets forgotten: without it, you would announce a rate of $1.21$ m/h, about twice the true value.

c)

$|h'(t)|$ is largest when $\sin\left(\frac{\pi t}{6.2}\right)=1$ , that is $t=3.1$ h, at mid-tide. The level is then falling at $\frac{2.5\pi}{6.2}\approx 1.27$ m/h. At the same moment, $h''(t)$ , proportional to $-\cos$ , is zero: this is an inflection point of the curve.

d)

Each interval lasts $\frac{6.2}{6}$ h, and the argument of the cosine advances by $\frac{\pi}{6}$ . First interval: $2.5\left(1-\cos\frac{\pi}{6}\right)\approx 0.33$ m, against $\frac{5}{12}\approx 0.42$ m for the rule. Third: $2.5\left(\cos\frac{\pi}{3}-\cos\frac{\pi}{2}\right)=1.25$ m, exactly $\frac{3}{12}\times 5$ . The rule is a rough approximation at the ends, where the level moves slowly, and accurate in the middle, where the derivative is largest. It translates into whole fractions what the derivative says: slow, fast, slow.

e)

$h(t)=2$ gives $\cos\left(\frac{\pi t}{6.2}\right)=-0.4$ , that is $\frac{\pi t}{6.2}=\arccos(-0.4)\approx 1.982$ and $t\approx 3.91$ h, about $3$ h $55$ min after high tide. At that moment, $h'\approx-1.267\times\sin(1.982)\approx-1.267\times 0.917\approx-1.16$ m/h: the level is still falling at almost full speed, and a few minutes of delay cost more than ten centimetres of water.

Exercise 15: Problem: the best viewing angle of a screen

A vertical screen has its bottom edge $1$ m above the eyes of a viewer and its top edge $4$ m above. The viewer stands at a horizontal distance $x>0$ from the wall, in metres. Let $\theta(x)$ be the angle under which the viewer sees the screen, in radians. The figure below shows the situation; it is not to scale.

a) Express $\theta(x)$ using two arctangents.
b) Compute $\theta'(x)$ and put it over a common denominator.
c) At what distance is the viewing angle largest? Justify with the sign of $\theta'$ and with the limits at $0^{+}$ and at $+\infty$ .
d) Give this maximum angle in degrees, then check that its tangent is exactly $\frac{3}{4}$ .
e) Show that, for a screen between heights $a$ and $b$ above the eyes, the best distance is $\sqrt{ab}$ . Apply this to a painting whose edges are $0.5$ m and $2$ m above the eyes.

Show the solution

Answers

a) $\theta(x)=\arctan\frac{4}{x}-\arctan\frac{1}{x}$
b) $\theta'(x)=\frac{12-3x^{2}}{(x^{2}+16)(x^{2}+1)}$
c) $x=2$ m
d) $36.87°$ , $\tan\theta=\frac{3}{4}$
e) $x=\sqrt{ab}$ ; $1$ m for the painting

a)

The line of sight to the top of the screen makes with the horizontal an angle whose tangent is $\frac{4}{x}$ , the line of sight to the bottom an angle whose tangent is $\frac{1}{x}$ . The screen is seen under their difference: $\theta(x)=\arctan\frac{4}{x}-\arctan\frac{1}{x}$ .

b)

$\left(\arctan\frac{k}{x}\right)'=\frac{-\frac{k}{x^{2}}}{1+\frac{k^{2}}{x^{2}}}=\frac{-k}{x^{2}+k^{2}}$ . So $\theta'(x)=\frac{-4}{x^{2}+16}+\frac{1}{x^{2}+1}=\frac{-4(x^{2}+1)+(x^{2}+16)}{(x^{2}+16)(x^{2}+1)}=\frac{12-3x^{2}}{(x^{2}+16)(x^{2}+1)}$ .

c)

The denominator is positive: $\theta'$ has the sign of $12-3x^{2}$ , positive for $x<2$ , negative for $x>2$ . The angle increases then decreases: maximum at $x=2$ m. At the ends, $\theta\to\frac{\pi}{2}-\frac{\pi}{2}=0$ as $x\to 0^{+}$ , since standing against the wall you see the screen edge-on, and $\theta\to 0-0=0$ as $x\to+\infty$ . The maximum is therefore absolute.

d)

$\theta(2)=\arctan 2-\arctan\frac{1}{2}\approx 63.43°-26.57°=36.87°$ . By the formula $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ : $\frac{2-0.5}{1+2\times 0.5}=\frac{1.5}{2}=\frac{3}{4}$ . The optimal angle is the one of the $3$ , $4$ , $5$ triangle.

e)

The same computation gives $\theta'(x)=\frac{-b}{x^{2}+b^{2}}+\frac{a}{x^{2}+a^{2}}$ , which is zero when $b(x^{2}+a^{2})=a(x^{2}+b^{2})$ , that is $(b-a)x^{2}=ab(b-a)$ and $x^{2}=ab$ : the best distance is the GEOMETRIC MEAN of the two heights. This gives back $\sqrt{1\times 4}=2$ . For the painting: $\sqrt{0.5\times 2}=1$ m. It is not the arithmetic mean $1.25$ : stepping back to the average of the heights would be too far.

Corrected exercises: derivatives of trigonometric, exponential and logarithmic functions (201-SN2-RE)

Course recap

Calculations: derivatives to compute

Exercise 0.1: Twelve derivatives to compute directly

Part A: the basics (/50)

Exercise 1: The table of the six trigonometric derivatives, rebuilt

Exercise 2: The exponential: what the base changes, and what it does not

Exercise 3: The logarithm, and the domain that comes before the derivative

Exercise 4: Nesting three layers without losing a factor

Exercise 5: Logarithmic differentiation: when no rule applies

Part B: problems and reasoning (/50)

Exercise 6: The inverse trigonometric functions and their algebraic derivatives

Exercise 7: The hyperbolic functions and the shape of the catenary

Exercise 8: Five statements to correct

Exercise 9: Problem: radioactive decay and the relative rate

Exercise 10: Problem: the logistic model and its maximum rate of spread

Part C: the classics (/50)

Exercise 11: The tangent lines to the exponential and the logarithm that pass through the origin

Exercise 12: Zero derivative, constant function: two identities of the inverse functions

Exercise 13: The horizontal tangent lines of e to the x times sin x

Exercise 14: Problem: the tide and the rule of twelfths

Exercise 15: Problem: the best viewing angle of a screen

See also

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