# The Mathematical Pirate's guide to the Chain Rule

“Yarr!” said the Mathematical Pirate, chopping his way through the piles of papers strewn around the classroom. The student looked unimpressed.

“How do you differentiate $y = \sec^2(3x)$?” she asked. The Mathematical Ninja would have discerned that she’d just read the question without trying to solve it, but the Mathematical Pirate didn’t value such psychology.

“Arr!” he said. “Differentiation is all about little bits!” He swigged from a bottle of rum. How he got through the CRB check, I’ve no idea.

“Little bits?” The student recognised that this was unconventional.

“Yarr. It’s a bit so little that if you square it, you get 0. A little bit of $x$ is $\delta x$. So, if you change $y$ by a little bit, you get $y + \delta y$. If you change $x$ by a little bit, you get $x + \delta x$.”

“OK, so if I change this equation, I get $y + \delta y = \sec^2(3x + 3\delta x)$?”

“Yarr, me hearty. Now you can use a more sensible function on the right - like $\cos$.”

“I think I see… $y + \delta y = \frac { 1 }{ (\cos(3x + 3 \delta x) )^2 }$.”

“Then expand the bottom with a compound angle formula…”

The student knew she could get away with looking that up; the Mathematical Pirate didn’t care about memory. “The bottom bracket would be $\cos(3x)\cos(3 \delta x) - \sin(3x)\sin(3\delta x)$.”

“Shiver me timbers! We need small-angle approximations! $\sin(z)$ is roughly $z$ for small angles in radians. $\cos(z)$ is roughly 1. And anything with a $\delta$ in is definitely a small angle.”

“Right… so $y + \delta y = \frac {1}{ (\cos(3x) + 3\sin(3x)\delta x)^2}$. Should I square the bottom? Of course I should square the bottom. It’s $\cos^2(3x) + 6\sin(3x)\cos(3x) \delta x + $ something with a $(\delta x)^2$ in, which is so small as to be 0.”

“Yarr. Nobody likes $\delta x$s on the bottom of a fraction, though. You should rationalise!”

It’s not irrational, thought the student, but he has a sword. And three sheets to the wind. “I’d use the conjugate trick from C1, and multiply top and bottom by… yikes… $\cos^2(3x) - 6 \sin(3x)\cos(3x) \delta(x)$. Oh… but that makes the bottom nicer! It’s just $\cos^4(3x)$, the second term goes away when you square it.”

“Hic!” said the Mathematical Pirate and muttered something under his breath about treasure.

The student carried on. “ $y + \delta y = \frac {\cos^2 (3x) - 6 \sin(3x)\cos(3x) \delta x}{\cos^4 (3x)}$ … that gets simpler, doesn’t it? $y + \delta y = \sec^2(3x) - 6 \frac{\sin(3x)}{\cos^3(3x)} \delta x$.”

“What shall we do wizh zhe drunken shailor,” said the Mathematical Pirate.

“For heaven’s sake, I’ll work it out myself. I can take away what I had to start with, to get $\delta y = - 6 \frac{\sin(3x)}{\cos^3(3x)} \delta x$… and that fraction looks like it’s $\tan(3x)\sec^2(3x)$… divide across and I have $\frac{\delta y}{\delta x} = -6 \tan(3x) \sec^2(3x)$. That looks like an answer!”

“Or you could have used the chain rule,” said the Mathematical Pirate, with an evil leery grin.

“Piece of cake,” said the Mathematical Parrot. “Piece of cake.”