Question

If $y=\left({\log }_{\cos x}\sin x\right)\left({\log }_{nx}\cos x\right)+{\sin }^{-1}\frac{2x}{1+x^2}$ then $\frac{dy}{dx}$ at $x=\frac{\pi }{2}$ is equal to

• A. $\frac{8}{(4+{\pi }^2)}$
• B. $0$
• C. $-\frac{8}{(4+{\pi }^2)}$
• D. $1$

Answer 1

The correct option is A $\frac{8}{(4+{\pi }^2)}$

$y=\frac{\log \sin x}{\log \cos x}.\frac{\log \cos x}{\log nx}+u={\sin }^{-1}\frac{2x}{1+x^2}$
$y=\frac{\log \sin x}{\log nx}+{\sin }^{-1}\frac{2x}{1+x^2}$
Let $u=\frac{\log \sin x}{\log nx},v=\frac{2x}{1+x^2}$
$\Rightarrow y=u+v{\left[\frac{dy}{dx}\right]}_{\pi /2}={\left[\frac{du}{dx}\right]}_{\pi /2}+{\left[\frac{dv}{dx}\right]}_{\pi /2}...\left(1\right)$
$u=\frac{\log \sin x}{\log nx}\Rightarrow \frac{du}{dx}=\frac{\left(\log nx\right)\cot x-\frac{\log \sin x}{x}}{{\left(\log nx\right)}^2}$
$\therefore {\left[\frac{du}{dx}\right]}_{x=\pi /2}=0$
$v={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
let $x=\tan \theta$
$\therefore v={\sin }^{-1}\left(\sin 2\theta \right)2\theta =2{\tan }^{-1}x$
$\therefore \frac{dv}{dx}=\frac{2x}{1+x^2}\therefore {\left[\frac{dv}{dx}\right]}_{x=\pi /2}=\frac{2}{1+\left({\pi }^2/4\right)}=\frac{8}{{\pi }^2+4}$