Dear Uncle Colin,

In a recent test, I was asked to differentiate $\frac{x^2+4}{\sqrt{x^2+4}}$. Obviously, my first thought was to simplify it to $\br{x^2+4}^{-\frac{1}{2}}$, but I’m not allowed to do that: only to use the quotient rule and the fact that $\diff {\sqrt{f(x)}}{x} = \frac{f’(x)}{2f(x)}$.

When Evaluating, Inappropriate Rules Demanded

Hi, WEIRD, and thanks for your message - that is indeed a weird restriction to put on this question! I would do it exactly the way you suggest. However, let’s play along.

### Without any simplification

Using the quotient rule, $\diff{\frac{u}{v}}{x}= \frac{v \diff ux - u \diff vx}{v^2}$, we have $u = x^2+4$ and $v = \sqrt{x^2 + 4}$.

Presumably, you’re allowed to differentiate $x^2+4$ normally to get $2x$; meanwhile, $\diff vx = \frac{2x}{2\sqrt{x^2+4}}$, using the rule you’re given (and those 2s cancel out).

Ignoring the $v^2$ on the bottom for now, the top is $v \diff ux - u \diff vx$, or $\br{\sqrt{x^2 + 4}}\br{2x} - \br{x^2+4}\br{\frac{x}{\sqrt{x^2+4}}}$.

Assuming we’re still not allowed to spot the simplification in the second term, let’s turn all of the top into a single fraction:

$\frac{ 2x\br{x^2+4} - x\br{x^2+4}}{\sqrt{x^2+4}}$, which can be simplified (presumably) to $\frac{ x\br{x^2+4}}{\sqrt{x^2+4}}$.

Now to tackle the $v^2$, which is just $x^2+4$.

This makes the full derivative $\frac{\br{\frac{ x\br{x^2+4}}{\sqrt{x^2+4}}}}{x^2+4}$.

Now, one would have to hope at this point you’re allowed to say “look, there’s an $x^2+4$ on the top and the bottom, and those cancel”, giving you $\frac{x}{\sqrt{x^2+4}}$.

### A little rant

I disagree, fundamentally, with saying “you must use this method” in pretty much any question. I’m fine with saying “you must use a valid method” or even “you must use a result you have proved”, but not “do it the hard way because I can’t be bothered to come up with a question that actually requires the rules I want to test.”

Grr. I feel your pain. I hope that helps.

- Uncle Colin