How do I go about solving this derivative?

Question

I have the function $$f(x)=\ln\sqrt{8+\cos^2x}$$ so $$1.f(x)=\ln(8+\cos^2x)^\frac{1}{2}$$so$$2.f(x)=\frac{1}{2}\ln(8+\cos^2x)$$so $$3.f'(x)=\frac{1}{2}\left[\frac{-2 \cos x^{\sin x}}{8+\cos ^2x}\right]$$so$$4.f'(x)=\frac{1}{2}\left[\frac{-\cos x^{\sin x}}{8+\cos ^2x}\right]$$ I'm having problems with everything after step 2, a step by step explanation would be greatly appreciated. Thanks. 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. !

Answer 1

Just use the straightforward differentiation rule: $$ f(x ) = \sqrt{\phi(x)}\\ f'(x) = \frac{1}{2}\frac{\phi'(x)}{\phi(x)^{\frac{3}{2}}} = \frac{1}{2}\frac{\frac{- \sin 2 x}{8+ \cos ^2 x}}{(\phi(x))^{\frac{3}{2}}} $$

EDIT: since the original function is quite different, then you're on the right way: pull out $\frac{1}{2}$ and use the chain rule. You'll get something like $-\frac{1}{2} \cdot \frac{\sin 2 x}{8 +\cos^2 x}$

Answer 2

Note the correction for $(3)$: $$f'(x)= \frac 12\cdot\frac{-2 \cos x\sin x}{8 + \cos^2 x}$$

$$= \dfrac{-\cos x\sin x}{8 + \cos^2 x}$$

$$=-\frac 12\cdot \frac{\sin(2x)}{8 + \cos^2 x}$$

I'm not at all clear as to why you raise $\cos x$ to the power of $\sin x$. You must have intended to write the numerator as you did, since I checked your formatting.