Continuing on from my collection of humorous mathematical aphorisms, today’s excerpt is an exercise from Calculus by Michael Spivak (4 ed.).

Chapter 10, Question 2. Find $f'(x)$ for each of the following functions $f$. (It took the author 20 minutes to compute the derivatives for the answer section, and it should not take you much longer: Although rapid calculation is not the goal of mathematics, if you hope to treat theoretical applications of the Chain Rule with aplomb, these concrete applications should be child’s play–mathematicians like to pretend that they can’t even add, but most of them can when they have to.)

i. $f(x)=\sin((x+1)^2(x+2))$

ii. $f(x)=\sin^3(x^2+\sin x)$

iii. $f(x) = \sin^2((x + \sin x)^2)$

iv. $f(x)=\sin\left( \frac{x^3}{\cos x^3}\right)$

v. $f(x)=\sin(x\sin x) + \sin(\sin x^2)$

vi. $f(x)=(\cos x)^{31^2}$

vii. $f(x)=\sin^2 x \sin x^2 \sin^2 x^2$

viii. $f(x)=\sin^3(\sin^2 (\sin x))$

ix. $f(x)=(x+\sin^5 x)^6$

x. $f(x)=\sin(\sin(\sin(\sin(\sin x))))$

xi. $f(x)=\sin((\sin^7 x^7 +1)^7)$

xii. $f(x)=(((x^2+x)^3+x)^4+x)^5$

xiii. $f(x)=\sin(x^2+\sin(x^2+\sin x^2))$

xiv. $f(x)=\sin(6\cos(6\sin(6\cos 6x)))$

xv. $f(x)=\frac{\sin x^2 \sin^2 x}{1+\sin x}$

xvi. $f(x)=\frac{1}{x-\frac{2}{x+\sin x}}$

xvii. $f(x)=\sin\left( \frac{x^3}{\sin\left( \frac{x^3}{\sin x} \right)} \right)$

xviii. $f(x)=\sin\left( \frac{x}{x-\sin\left( \frac{x}{x-\sin x}\right)} \right)$

I did the first 8 questions in ~8 minutes before the tedium got to me, so I’m going to assume Spivak’s estimate is at least on the correct order of magnitude.

Credit: Leo Fouchè for sending this one through.