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

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    $\begingroup$ 5th line it should be $\cos^5x$ $\endgroup$ – Yadati Kiran Nov 26 '18 at 16:03
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    $\begingroup$ last line when you factor $\sin x$ you have $2\sin^2x-\cos^2x+\cos^4x$. $\endgroup$ – Yadati Kiran Nov 26 '18 at 16:06
  • $\begingroup$ How did you get to that? I got $2\sin^2x+\cos^2x+\cos^4x$ $\endgroup$ – Aleksa Nov 26 '18 at 16:10
  • $\begingroup$ Ah yes I see, I made a mistake in the last step $\endgroup$ – Aleksa Nov 26 '18 at 16:12

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


Note after fixing sine power mistake in the last step you have $$ \cos^4 x - \cos^2 x + 2\sin ^2 x = \cos^4 x - 3\cos^2 x + 2 = \left(\cos^2 x - 1\right)\left(\cos^2 x - 2\right) $$

  • $\begingroup$ Yes I see just changed it now, before I saw your answer $\endgroup$ – Aleksa Nov 26 '18 at 16:04
  • $\begingroup$ Even after fixing it I still don't know what to do $\endgroup$ – Aleksa Nov 26 '18 at 16:06
  • $\begingroup$ @Aleksa see update $\endgroup$ – gt6989b Nov 26 '18 at 16:10
  • $\begingroup$ I just checked the results of this from the textbook, the solution is $\tan^5x$, so there is more to do I think $\endgroup$ – Aleksa Nov 26 '18 at 16:14
  • $\begingroup$ I got the results, the problem was $1\over4$ instead of ${1\over2}$ made a typo while writing this down and doing it on the computer $\endgroup$ – Aleksa Nov 26 '18 at 16:21

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

  • $\begingroup$ Is there something else to do after that point? The solution from the textbook is $\tan^5x$ $\endgroup$ – Aleksa Nov 26 '18 at 16:19
  • $\begingroup$ I edited my post, there was a typo at the beginning, got the solution after changing everything $\endgroup$ – Aleksa Nov 26 '18 at 16:23
  • $\begingroup$ Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last. $\endgroup$ – Yadati Kiran Nov 26 '18 at 16:23

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