Differentiate $x^{\sin x}, x > 0$ w.r.t. $x$.

$x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \ln x \right]$

The correct answer is Option (3) → $x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \ln x \right]$ ##

Let $y = x^{\sin x}$. Taking logarithm on both sides, we have

$\log y = \sin x \log x$

Therefore $\frac{1}{y} \frac{dy}{dx} = \sin x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(\sin x)$

or $\frac{1}{y} \frac{dy}{dx} = (\sin x) \frac{1}{x} + \log x \cos x$

or $\frac{dy}{dx} = y \left[ \frac{\sin x}{x} + \cos x \log x \right]$

$= x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right]$