1. ## Derivative check

The question is:

$y= (lntanx)^2$

I did:
$
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.

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

3. 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?

4. 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)).

5. Alright, good to know. Thank you Halls.