If $y=\sin^{-1}(\sin x)$, then $\frac{dy}{dx}$ at $x=\frac{π}{2}$ is:
Answer & explanation
Correct answer: option 3
We have, $y=\sin^{-1}(\sin x)$
$y=\left\{\begin{matrix}x,&\frac{-\pi}{2}≤x≤\frac{\pi}{2}\\\pi-x,&\frac{\pi}{2}≤x≤\pi\end{matrix}\right.⇒\frac{dy}{dx}=\left\{\begin{matrix}1,&\frac{-\pi}{2}≤x≤\frac{\pi}{2}\\dose\,not\,exist,&x=\frac{\pi}{2}\\-1,&x>\frac{\pi}{2}\end{matrix}\right.$