
Derivative check
The question is:
$\displaystyle y= (lntanx)^2$
I did:
$\displaystyle
y'= 2(lntanx) * [ \frac{1}{x} + lnsec^2x]$
Out of all that I've done it seems like it's more along the easy side, but you can never be too sure. Just want to double check my work. Thanks.

$\displaystyle 2\ln(\tan(x))\cdot\frac{\sec^2(x)}{\tan(x)}=\frac{ 2\ln(\tan(x))\sec^2(x)}{\tan(x)}=\cdots$

I took it as a product rule...it was written exactly as I typed, not as ln(tan(x)). If it was like that, I'd have done 1/tanx * sec^2(x), just as you had. But, since it wasn't written in that way, I got thrown off. Am I missing something here?

You are missing the fact that (ln)(tan x) as a product is meaningless there is no "x" in the logarithm. It can only mean ln(tan(x)).

Alright, good to know. Thank you Halls.