I have the function

$$\tag{1} f(x)=\ln\sqrt{8+\cos^2x}$$

So we derive it as follows:

$$\tag{2} f(x)=\ln(8+\cos^2x)^\frac{1}{2}$$

$$\tag{3} f(x)=\frac{1}{2}\ln(8+\cos^2x)$$

$$\tag{4} f'(x)=\frac{1}{2}\left[\frac{-2 \cos x{\sin x}}{8+\cos^2x}\right]$$

$$\tag{5} f'(x)=\left[\frac{-\cos x{\sin x}}{8+\cos ^2x}\right]$$

I'm having problems with everything after step 2, a step by step explanation would be greatly appreciated. Thanks. For example where does the sin come from and why is there a cos in the numerator now. A detailed process of beginning to end of solving this (what laws are used and why etc..) would be greatly appreciated.

Here is the image of the website and how it goes about doing the steps, I thought it's be better for me to try and type it out unless anybody is curious. From the site. !

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    $\begingroup$ It's a straightforward application of the chain rule: $\frac{d}{dx}\ln{f(x)}=\frac{\frac{d}{dx}f(x)}{f(x)}$. $\endgroup$ – David H Jun 29 '14 at 1:19

Recall that $\frac{d}{dx}\ln(x) = \frac1x$. So using the chain rule, $$\frac{d}{dx}\ln(f(x)) = \frac{f'(x)}{f(x)}$$

Do you think you can work it out from there?

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    $\begingroup$ I'm about to cry at how easy this problem was and how much I over-thinked it haha! Thanks for "opening my eyes". $\endgroup$ – Kenshin Jun 29 '14 at 1:41
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    $\begingroup$ Happens to all of us :( $\endgroup$ – Mathmo123 Jun 29 '14 at 11:55

You just have to use the Chain Rule a couple of times.

The derivative of $$(1/2)\ln(8+\cos^2 x)$$

is $$(1/2)(1/(8+\cos^2 x))(2\cos x(-\sin x)).$$


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