Practicing Success
If $y=\left(\log _{\cos x} \sin x\right)\left(\log _{\sin x} \cos x\right)+\sin ^{-1} \frac{2 x}{1+x^2}$, then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is equal to |
$\frac{8}{\pi^2+4}$ 0 $\frac{-8}{\pi^2+4}$ 1 |
$\frac{-8}{\pi^2+4}$ |
We have, $y=\left(\log _{\cos x} \sin x\right) \times\left(\log _{\sin x} \cos x\right)+\sin ^{-1} \frac{2 x}{1+x^2}$ $\Rightarrow y= \begin{cases}1-\pi-2 \tan ^{-1} x, & x<-1 \\ 1+2 \tan ^{-1} x, & \text { if }-1 \leq x \leq 1 \\ 1+\pi-2 \tan ^{-1} x, & \text { if } x>1\end{cases}$ $\Rightarrow \frac{d y}{d x}=\left\{\begin{array}{l}\frac{-2}{1+x^2}, x<-1 \text { or } x>1 \\ \frac{2}{1+x^2},-1<x<1\end{array}\right.$ $\Rightarrow \left(\frac{d y}{d x}\right)_{x=\pi / 2}=\frac{-2}{1+\frac{\pi^2}{4}}=\frac{-8}{4+\pi^2}$ |