Solve this

We are given with an equation $y=\left\{\log _{\cos x} \sin x\right\}\left\{\log _{\sin x} \cos x\right\}^{-1}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, we have to find $\frac{d y}{d x}$ at

$x=\frac{\pi}{4}$ by using the given equation, so by differentiating the equation on both sides with respect to $x$, we get,

By using the properties of logarithms,

$y=\left\{\log _{\cos x} \sin x\right\}^{2}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

$y=\left\{\frac{\ln \sin x}{\ln \cos x}\right\}^{2}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

$\frac{d y}{d x}=2\left\{\frac{\ln \sin x}{\ln \cos x}\right\} \frac{\ln \cos x \frac{\cos x}{\sin x}-\ln \sin x \frac{-\sin x}{\cos x}}{(\ln \cos x)^{2}}+\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}} \frac{\left(1+x^{2}\right) 2-2 x(2 x)}{\left(1+x^{2}\right)^{2}}$

$\frac{d y}{d x}=2\left\{\frac{\ln \sin x}{\ln \cos x}\right\} \frac{\ln \cos x(\cot x)-\ln \sin x(-\tan x)}{(\ln \cos x)^{2}}+\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{\left(1-x^{2}\right)^{2}}} \frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$

$\frac{d y}{d x}=2\left\{\frac{\ln \sin x}{\ln \cos x}\right\} \frac{\ln \cos x(\cot x)+\ln \sin x(\tan x)}{(\ln \cos x)^{2}}+\frac{2}{1+x^{2}}$

Now putting the value of $x=\frac{\pi}{4}$ in the derivative solved above, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{x}=\pi / 4)=2\{1\} \frac{\ln \frac{1}{\sqrt{2}}(1)+\ln \frac{1}{\sqrt{2}}(1)}{\left(\ln \frac{1}{\sqrt{2}}\right)^{2}}+\frac{2}{1+\left(\frac{\pi}{4}\right)^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{x}=\mathrm{r} / 4)=2\{1\} \frac{\ln \frac{1}{2}}{\left(\frac{1}{2} \ln 2\right)^{2}}+\frac{2}{\frac{16+\pi^{2}}{16}}$

$\frac{d y}{d x}(x=\pi / 4)=2\{1\} \frac{-4 \ln 2}{(\ln 2)^{2}}+\frac{32}{16+(\pi)^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{x}=\pi / 4)=\frac{-8}{\ln 2}+\frac{32}{16+(\pi)^{2}}$