Question

If $y={\tan }^{-1}\sqrt{\frac{\sin x}{1+\cos x}}$ then find $\frac{dy}{dx}$

Answer 1

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

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

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

$=\frac{1}{\sin x+\cos x+1}\cdot \frac{1}{\sqrt{\sin x(\cos x+1)}}$