<s>[INST] 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.
 [/INST] 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 $x\ne 0$.
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