Question

Find $\frac{dy}{dx}$ for $y={\sin }^{-1}\left(\cos x\right)$ , where $x\in \left(0,2\pi \right)$

Answer 1

$y={\sin }^{-1}\left(\cos x\right)$

$\sin y=\cos x$

$\cos y=\sqrt{1-{\cos }^{2}x}-----\left(1\right)$

$\cos \left(\frac{dy}{dx}\right)=-\sin x$

$\frac{dy}{dx}=\frac{1}{\cos y}(-\sin x)=\frac{-\sin x}{\sqrt{1-{\cos }^{2}x}}$ (from $\left(1\right)$)

$\therefore \frac{dy}{dx}=\frac{-\sin x}{\sqrt{1-{\cos }^{2}x}}=-1\therefore \frac{dy}{dx}=-1$