Question

Find the derivative of ${\cos }^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)$ , $\frac{-\pi }{4}<x<\frac{\pi }{4}$.

Answer 1

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

$={\cos }^{-1}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)$

$={\cos }^{-1}(\sin \frac{\pi }{4}\sin x+\cos \frac{\pi }{4}\cos x)$

$={\cos }^{-1}(\cos \frac{\pi }{4}\cos x+\sin \frac{\pi }{4}\sin x)$

$={\cos }^{-1}\left(\cos (\frac{\pi }{4}-x)\right)$

$y=\frac{\pi }{4}-x$

$\frac{dy}{dx}=-1$