Question

If $y={cos}^{-1}\left(\frac{sinx+cosx}{\sqrt{2}}\right)$ where $\frac{\pi }{4}<x<\frac{\pi }{4}$ then find $\frac{dy}{dx}$

Answer 1

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

$\Rightarrow \cos y=\frac{\sin x+\cos x}{\sqrt{2}}$

$\Rightarrow \cos y=\cos x\cos \frac{\pi }{4}+\sin x\sin \frac{\pi }{4}$

$\Rightarrow \cos y=\cos (x-\frac{\pi }{4})$

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

$\Rightarrow \frac{dy}{dx}=1-0=1$

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