"Arr, that be a scurvy-lookin' expression!" said the Mathematical Pirate. "A quartic on the top and a quadratic on the bottom. That Ninja would probably try to factorise and do it all elegant-like."

"Is that not the point?"

"When you got something as 'orrible as that, it's like puttin' lipstick on a shark. There ain't all that much point, and it's just gonna annoy the shark."

The student looked both ways. It was not a metaphor he was familiar with. "Right you are, then, captain. So we've got to find the gradient of $f(x) = \frac{x^4 + x^3 - 13x^2 + 26x - 17}{x^2 - 3x + 3}$ when $x=1$. That's looking all quotient-ruley to me."

"Arr," nodded the Mathematical Pirate.

"I mean, $u$ and $v$ on their own look ok, and $u'$ and $v'$ are even nicer, but their products..."

"Agreed!" agreed the Mathematical Pirate. "Work them out anyway!"

"Ok. $u = x^4 + x^3 - 13x^2 + 26x - 17,$, so $u' = 4x^3 + 3x^2 - 26x + 26$; $v = x^2 - 3x + 3$, so $v' = 2x - 3$."

"Now figure those out when $x=1$."

The student's eyes opened wide. "You mean..."

"I mean. You don't have to do all that algebra to find the gradient - you can just put a value in and use the quotient rule on the result. Works for the other rules, too."

The student looked hurt, mildly offended, and then overjoyed.

"So, $u = -2$, $u' = 7$, $v = 1$ and $v' = -1$. That's already a lot nicer. $\frac{vu' - uv'}{v^2} = \frac{7 - 2}{1} = 5 $. Is that it?!"

"Ahar," said the Mathematical Pirate. "Don't tell the Ninja."

* Edited 2020-01-16 because I'd said "Ninja" instead of "Pirate" and now I fear for my life. Thanks a bunch, Adam.