Question

Derivative of $\log \sin x$ w.r.t. $\sqrt{\cos x}$ is

• A. $-2\sqrt{\cos x}\cot x\csc x$
• B. $+2\sqrt{\cos x}\cot x\csc x$
• C. $-2\sqrt{\cot x}\cos x\csc x$
• D. None of these

Answer 1

The correct option is A $-2\sqrt{\cos x}\cot x\csc x$

Let $u=\log \sin x$ and $v=\sqrt{\cos x}$

Differentiate w.r.t. $x$

$\frac{du}{dx}=\frac{d}{dx}\left(\log \sin x\right)$ and $\frac{dv}{dx}=\frac{d}{dx}\sqrt{\cos x}$

$\frac{du}{dx}=\frac{1}{\sin x}\frac{d}{dx}\left(\sin x\right)$and $\frac{dv}{dx}=\frac{1}{2\sqrt{\cos x}}\frac{d}{dx}\left(\cos x\right)$

$\frac{du}{dx}=\cot x$ and $\frac{dv}{dx}=-\frac{\sin x}{2\sqrt{\cos x}}$

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\cot x}{-\frac{\sin x}{2\sqrt{\cos x}}}$

$=-2\sqrt{\cos x}\cot x\csc x$