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{d y}{d x}=\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$
$d t / d x=-\sin x$
$ \text { At } x=\frac{\pi}{2}, \frac{d t}{d x}=-1$
$ \therefore \frac{d y}{d x}=\frac{\left(\frac{2}{\pi}\right)}{-1}=\frac{-2}{\pi} $