Question

Differentiate w.r.t. $x$ the function: $cot^{-1}\bigg[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\bigg],0<x<\frac{x}{2}$

Answer 1

The correct answer is $\frac{dy}{dx}=\frac{1}{2}$
Let $y=cot^{-1}\bigg[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\bigg]......(1)$
Then,$\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}$
$=\frac{(\sqrt{1+sinx}+\sqrt{1-sinx)^2}}{(\sqrt{1+sinx}-\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}$
$=\frac{(1+sinx)+(1-sinx)+2\sqrt{(1-sinx)(1+sinx)}}{(1+sinx)-(1-sinx)}$
$=\frac{2+2\sqrt{1-sin^2x}}{2sinx}$
$=\frac{1+cosx}{sinx}$
$=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}$
$=cot\frac{x}{2}$
$y=cot^{-1}(cot\frac{x}{2})$
$⇒y=\frac{x}{2}$
$∴\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}(x)$
$⇒\frac{dy}{dx}=\frac{1}{2}$