Question

If $y={\tan }^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}$, find $\frac{dy}{dx}$.

Answer 1

$y={\tan }^{-1}\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$

$y={\tan }^{-1}1$ since $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=1$

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

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{4}\right)=0$