Question

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

Answer 1

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