Question

Differentiate the given function w.r.t. $x$:
${\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}\sqrt{1-\sin x}}\right]$, 0 < x <$\frac{\pi }{2}$

Answer 1

Let $y={\cot }^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$ ...(i)

Then, $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

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

$=\frac{(1+\sin x)+(1-\sin x)+2\sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)-(1-\sin x)}$
$=\frac{2+2\sqrt{1-{\sin }^{2}x}}{2\sin x}$
$=\frac{1+\cos x}{\sin x}=\frac{2{\cos }^2\frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\cot \frac{x}{2}$
Therefore equation (1) becomes,
$y={\cot }^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}$
$\therefore \frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left(x\right)=\frac{1}{2}$