Differentiate the function $(\sin x)^{\cos x}$ with respect to $x$.

$\sin x^{\cos x} \left[ \frac{\cos^2 x}{\sin x} - \sin x \cdot \log (\sin x) \right]$

The correct answer is Option (2) → $\sin x^{\cos x} \left[ \frac{\cos^2 x}{\sin x} - \sin x \cdot \log (\sin x) \right]$ ##

Let $y = (\sin x)^{\cos x}$

Taking log on both sides, we get

$\log y = \log(\sin x)^{\cos x} = \cos x \log \sin x \quad [∵\log m^n = n \log m]$

On differentiating w.r.t. $x$, we get

$\frac{d}{dy} \log y \cdot \frac{dy}{dx} = \frac{d}{dx} (\cos x \cdot \log \sin x)$

$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \frac{d}{dx} \log \sin x + \log \sin x \cdot \frac{d}{dx} \cos x$

$\left[ ∵\frac{d}{dx} [f(x) \cdot g(x)] = f(x) \frac{d}{dx} g(x) + g(x) \frac{d}{dx} f(x) \right]$

$= \cos x \cdot \frac{1}{\sin x} \frac{d}{dx} \sin x + \log \sin x \cdot (-\sin x)$

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

$∴\frac{dy}{dx} = y \left[ \frac{\cos^2 x}{\sin x} - \sin x \cdot \log (\sin x) \right]$

$= \sin x^{\cos x} \left[ \frac{\cos^2 x}{\sin x} - \sin x \cdot \log (\sin x) \right] \quad [∵y = (\sin x)^{\cos x}]$