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

Answer: (A)

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