Find the derivative of $y={1\over2}\tan^4x-{1\over2}\tan^2x-\ln\cos x$

Question

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

Answer 1

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 $$ \cos^4 x - \cos^2 x + 2\sin ^2 x = \cos^4 x - 3\cos^2 x + 2 = \left(\cos^2 x - 1\right)\left(\cos^2 x - 2\right) $$

Answer 2

$\displaystyle{\sin x(2\sin^2x-\cos^2x+\cos^4x)\over\cos^5x}={\sin x(2\sin^2x-\cos^2x(1-\cos^2x))\over\cos^5x}$ $\displaystyle={\sin^4 x(1+\sin^2x)\over\cos^5x}=\tan^4x(\sec x+\tan x\sin x)$