Question

Find the derivative with respect to $x$ of the function $y=\left\{\left({\log }_{\cos x}\sin x\right){\left({\log }_{\sin x}\cos x\right)}^{-1}+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right\}$ at $x=\frac{\pi }{4}$.

Answer 1

Given, $y=\left\{\left({\log }_{\cos x}\sin x\right){\left({\log }_{\sin x}\cos x\right)}^{-1}+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right\}$
$\therefore y={\left(\frac{{\log }_e(\sin x)}{{\log }_e(\cos x)}\right)}^2+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$\Rightarrow \frac{dy}{dx}=2\left\{\frac{{\log }_e(\sin x)}{{\log }_e(\cos x)}+\frac{\left({\log }_e(\cos x).\cot x{\log }_e(\sin x)\cdot \tan x\right)}{{\left\{{\log }_e(\cos x)\right\}}^2}\right\}+\frac{2}{1+x^2}$
$\Rightarrow \left(\frac{dy}{dx}\right){\left(x=\frac{\pi }{4}\right)}^{=2}\left\{1\cdot \frac{2\cdot \log \left(\frac{1}{\sqrt{2}}\right)}{{\left(\log \frac{1}{\sqrt{2}}\right)}^2}\right\}+\frac{2}{1+\frac{{\pi }^2}{16}}$
$=-\frac{8}{\log e^2}+\frac{32}{16+{\pi }^2}$