# Ask Uncle Colin: Simplify *this*!

Dear Uncle Colin,

I’ve differentiated using the quotient rule and come up with $\frac{\frac{14x}{7x^2+6}x^7 -7x^6\ln\br{7x^2+6}}{\br{x^7}^2}$. That’s… not as simple as I’d like it to be. Any hints?

Quite Ugly Outcome That I’d Expect Neater, Thanks

Hi, QUOTIENT, and thanks for your message!

You’re right, that’s Not Very Nice. Let’s break it down a bit at a time.

The first question I would ask is, “what’s ugly?”. The second question is “no, what *specifically*?”, because the first answer is always “… the whole thing.”

For me, the ugliest thing is the stacked fraction, so I’d start by multiplying top and bottom by $7x^2 + 6$ to get:

$\frac{14x^8 - 7x^6\br{7x^2 + 6}\ln\br{7x^2 + 6}}{x^{14}\br{7x^2+6}}$

You’ll notice I’ve tidied up a few powers there as well. There’s also a factor of $x^6$ everywhere I can take out on top and on the bottom (assuming $x\ne0$):

$\frac{14x^2 - 7\br{7x^2 + 6}\ln\br{7x^2+6}}{x^8\br{7x^2 + 6}}$

There’s not a whole lot more to do, other than take out the factor of 7 on top to get:

$\frac{7\br{x^2 - \br{7x^2 + 6}}\ln\br{7x^2+6}}{x^8\br{7x^2 + 6}}$

I think that’s in its most useful form, although you *could* split it into two fractions as

$\frac{14}{x^6\br{7x^2+6}} - \frac{7\ln{7x^2+6}}{x^8}$.

Hope that helps!

- Uncle Colin