Question

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

• A. $8(\frac{4}{{\pi }^2+16}-\frac{1}{\log 2})$
• B. $-8(\frac{4}{(\pi +4{)}^2}-\frac{1}{\log 2})$
• C. $8(\frac{4}{{\pi }^2+16}+\frac{1}{\log 2})$
• D. $8(\frac{4}{(\pi +4{)}^2}-\frac{1}{\log 2})$

Answer 1

The correct option is A $8(\frac{4}{{\pi }^2+16}-\frac{1}{\log 2})$

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

$=({\log }_{\cos x}\sin x{)}^2+2{\tan }^{-1}x[\because {\log }_{b}a=\frac{1}{{\log }_{a}b}]$ $={\left(\frac{{\log }_e\sin x}{{\log }_e\cos x}\right)}^2+2{\tan }^{-1}x.$

$\frac{dy}{dx}=2\left(\frac{\log \sin x}{\log \cos x}\right)\cdot \frac{\cot x.\log \cos x+\tan x\log \sin x}{(\log \cos x{)}^2}+\frac{2}{1+x^2}$

Hence at $x=\frac{\pi }{4}.$

we have $\frac{dy}{dx}=2.\frac{\log \left(1/\sqrt{2}\right)}{\log \left(1/\sqrt{2}\right)}\frac{1.\log \left(1/\sqrt{2}\right)+1.\log \left(1/\sqrt{2}\right)}{\left[\log \right(1/\sqrt{2}){]}^2}+\frac{2}{1+\left({\pi }^2/16\right)}=\frac{-8}{\log 2}+\frac{32}{{\pi }^2+16}$

$=8(\frac{4}{{\pi }^2+16}-\frac{1}{\log 2}).$