Question

The derivate of $(\log x{)}^{\sin x}$ with respect to $\cos x$ at $x=\frac{\pi }{2}$ is

• A. $\frac{-4}{\pi }$
• B. $\frac{-\pi }{2}$
• C. $\frac{-2}{\pi }$
• D. $\frac{-\pi }{4}$

Answer 1

Answer: (C)

$y=(\log x{)}^{\sin x}$
$\log y=\sin x\cdot \log (\log x)$
$\frac{1}{y}\frac{dy}{dx}=\sin x\cdot \left(\frac{1}{\log x}\right)\left(\frac{1}{x}\right)+\log (\log x)\cos x$
$\text{ At }x=\frac{\pi }{2},\left[\frac{1}{\log \left(\frac{\pi }{2}\right)}\right]y^{\prime }=\frac{2}{\pi }\left[\frac{1}{\log \left(\frac{\pi }{2}\right)}\right]=\frac{2}{\pi }\text{ Let }t=\cos x$
$dt/dx=-\sin x$
$\text{ At }x=\frac{\pi }{2},\frac{dt}{dx}=-1$
$\therefore \frac{dy}{dx}=\frac{\left(\frac{2}{\pi }\right)}{-1}=\frac{-2}{\pi }$