Differentiate $f ( x ) = \frac { \ln \left( x ^ { 2 } \cos ( x ) \right) } { \sqrt { 1 - x ^ { 2 } } }$ could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.
 A: I would use logarithmic differentiation. Setting $u(x)=x^2\cos x$, we have
\begin{align}
\frac{f'(x)}{f(x)}&=\frac{u'(x)}{u(x)\ln u(x)}-\frac12\frac{-2x}{1-x^2} =\frac{2x\cos x-x^2\sin x}{x^2\cos x\,\ln(x^2\cos x)}-\frac12\frac{-2x}{1-x^2}\\[1ex]
&=\frac{2\cos x-x\sin x}{x\cos x\,\ln(x^2\cos x)}+\frac x{1-x^2},
\end{align}
whence
$$f'(x)=\frac{f'(x)}{f(x)}\,f(x)=\frac{2\cos x-x\sin x}{x\cos x\sqrt{1-x^2}}+\frac{x\ln(x^2\cos x)}{(1-x^2)^{3/2}}.$$
A: The quotient rule is a good idea. First differentiate $\ln(x^2 \cos (x))$ with the chain rule and the product rule.
A: 
I thought about the quotient rule 

That's right!
There's a quotient of the functions $g(x)= \ln \left( x ^ { 2 } \cos ( x ) \right)$ and $h(x)= \sqrt { 1 - x ^ { 2 } }$ so if you can find $g'(x)$ and $h'(x)$, the derivative $f'(x)$ follows as:
$$f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$$
I don't know why you think you'd need another quotient rule afterwards?
For $g'(x)$ and $h'(x)$ you need to apply the chain rule and for the "inner part" $x ^ { 2 } \cos ( x )$ you'll need the product rule as well: break it down step by step.
A: We have $${d\over dx}\ln {x^2\cos x}={1\over x^2\cos x}\cdot {(2x\cos x-x^2\sin x)}\\{d\over dx}\sqrt{1-x^2}=-{x\over \sqrt{1-x^2}}$$therefore by defining $g(x)=\ln x^2\cos x$ and $h(x)=\sqrt{1-x^2}$ and using $\left({g\over h}\right)'={g'h-gh'\over h^2}$ we finally obtain$$f'(x){=\left({g\over h}\right)'=\left({g'h-gh'\over h^2}\right)(x)\\={\left({1\over x^2\cos x}\cdot {(2x\cos x-x^2\sin x)}\right)\sqrt{1-x^2}+\left({x\over \sqrt{1-x^2}}\right)\ln x^2\cos x\over 1-x^2}\\={\left({2\over x} {-\tan  x}\right)\sqrt{1-x^2}+\left({x\over \sqrt{1-x^2}}\right)\ln x^2\cos x\over 1-x^2}\\={\left({2} {-x\tan  x}\right)(1-x^2)+x^2\ln x^2\cos x\over x(1-x^2)\sqrt{1-x^2}}}$$ for $|x|<1$ and $x\ne 0$.
