If $y=\sin^{-1}(\cos x)+\cos^{-1}(\sin x),0<x<\frac{π}{2}$, then $\frac{dy}{dx}$ is:
Answer & explanation
Correct answer: option 1
$y=\sin^{-1}(\cos x)+\cos^{-1}(\sin x)=(\frac{π}{2}-x)+(\frac{π}{2}-x)=π-2x$
$\frac{dy}{dx}=-2$
If $y=\sin^{-1}(\cos x)+\cos^{-1}(\sin x),0<x<\frac{π}{2}$, then $\frac{dy}{dx}$ is:
Correct answer: option 1
$y=\sin^{-1}(\cos x)+\cos^{-1}(\sin x)=(\frac{π}{2}-x)+(\frac{π}{2}-x)=π-2x$
$\frac{dy}{dx}=-2$