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Maybe the first step would be to change it to y = x(cos(x))/sin(x) and then take the derivative?

\[y'=\frac{\tan x(1)-x(\sec ^2x)}{\tan ^2x}=

...what?

Well it didn't post right. I'll try it again.

lol, yea I figured so :P

\[y'=\frac{\tan x-x(\sec ^2x)}{\tan ^2x}\]

Ok, so is that the answer?

\[\frac{\frac{\sin x}{\cos x}-\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}\]

Oh, I see

is that all y'?

Which is:
\[\frac{\cos x}{\sin x}-x \csc ^2x=\cot x-x \csc ^2x=y'\]

Is that it?