Differentiate $\frac{x}{\sin x}$ w.r.t. $\sin x$.

$\frac{\sin x - x \cos x}{\sin^2 x \cos x}$

The correct answer is Option (2) → $\frac{\sin x - x \cos x}{\sin^2 x \cos x}$ ##

Let $u = \frac{x}{\sin x}$ and $v = \sin x$

$∴\frac{du}{dx} = \frac{\sin x \cdot \frac{d}{dx} x - x \cdot \frac{d}{dx} \sin x}{(\sin x)^2} \quad \text{[by quotient rule]}$

$= \frac{\sin x - x \cos x}{\sin^2 x} \quad \dots (i)$

and $\frac{dv}{dx} = \frac{d}{dx} \sin x = \cos x \quad \dots (ii)$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\sin x - x \cos x) / \sin^2 x}{\cos x}$

$= \frac{\sin x - x \cos x}{\sin^2 x \cos x} = \frac{\frac{\sin x - x \cos x}{\cos x}}{\frac{\sin^2 x \cos x}{\cos x}}$ $\text{[dividing by } \cos x \text{ in both numerator and denominator]}$

$=\frac{\sin x - x \cos x}{\sin^2 x \cos x}$