<s>[INST] Find the derivative of $y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$ 
What is the derivative of 
  $$y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$$

What I did here was:
$$\begin{align}y'&=({1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x)'\\&={1\over2}\cdot4\tan^3x(\tan x)'-{1\over2}\cdot2\tan{x}(\tan x)'-{1\over\cos x}(\cos{x})'\\&={2\tan^3x\over\cos^2x}-{\tan x\over\cos^2x}+{\sin x\over\cos x}\\&={2{\sin^3x\over\cos^3x}\over\cos^2x}-{{\sin x\over\cos x}\over\cos^2x}+{\sin x\over\cos x}\\&={2\sin^3x\over\cos^5x}-{\sin x\over\cos^3x}+{\sin x\over\cos x}\\&={2\sin^3x-\sin x\cos^2x+\sin x\cos^4x\over\cos^5x}\\&={\sin x(2-\cos^2x+\cos^4x)\over\cos^5x}\end{align}$$
I'm not sure if this is the end, could this be further simplified in some way? And is it correct to this point at all?
I made a mistake in the beginning while writing down this problem, it should be ${1\over4}\tan^4x$, I changed it out and got exact results
 [/INST] You made a mistake, the first denominator should be $\cos^2 x \times \cos^3 x = \cos^5 x$ and you should be able to do more factoring then
UPDATE
Note after fixing sine power mistake in the last step you have
$$
\cos^4 x - \cos^2 x + 2\sin ^2 x
= \cos^4 x - 3\cos^2 x + 2
= \left(\cos^2 x - 1\right)\left(\cos^2 x - 2\right)
$$
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