Differentiate $\ln(\cos2x)$ With respect to $x$. I need to differentiate $\,\ln(\cos2x)$. 
Can someone please explain how to do this question? Thank you.
 A: Hints:


*

*What is the derivative of $\ln u = \dfrac{1}{u}$

*What is the derivative of $\cos w = -\sin w$

*What is the derivative of $2x = 2$?


Can you put these together?
Okay, from the comments and the answers above, we have:
$$\dfrac{d}{dx} \ln \cos 2x = \dfrac{1}{\cos 2x} (-\sin 2x) (2) = -2 \tan 2x$$
Do you see how we applied the chain rule?
A: $$\frac d{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$$
$$f(x) = \cos 2x$$
$$f'(x) = (2x)'(-\sin 2x) = -2\sin 2x $$
$$\frac d{dx} \ln(\cos 2x) = \dfrac {-2\sin 2x}{\cos 2x} = -2\tan 2x$$
A: Just use the chain rule. 
The chain rule says that $g(f(x)) = f'(x)\cdot g'(f(x))$
Now, here $f(x)=\cos(2x)$ and $g(f(x))=\ln(\cos(2x))$.
Remark: $g'(f(x))$ means that first you form $g'(u)$ and then you plug in $f(x)$ like you're evaluating $g'(u)$ at $u=f(x)$. 
More precisely, $g'(f(x))= \displaystyle {1 \over \cos(2x)}$ and as you already know $f'(x)= -2 \sin(2x)$. Now all you need to do is use chain rule. Then it becomes:
$g(f(x)) = f'(x)\cdot g'(f(x)) = -2 \sin(2x) \cdot \displaystyle {1 \over \cos(2x)} = -2 \tan(2x)$
A: Hint : 
$$\frac{df}{dx}=\frac{df}{du}\frac{du}{dt}\frac{dt}{dx}$$
put $f=\ln(\cos(2x))$ , $u=\cos(2x)$,$t=2x$
then $$\frac{d\ln(\cos(2x))}{dx}=\frac{d\ln(\cos(2x))}{d\cos(2x)}\frac{d\cos(2x)}{d2x}\frac{d2x}{dx}$$
