Differentiate $\sin^2 x$ w.r.t. $e^{\cos x}$.

$-2 \cos x e^{-\cos x}$

The correct answer is Option (3) → $-2 \cos x e^{-\cos x}$ ##

Let $u(x) = \sin^2 x$ and $v(x) = e^{\cos x}$. We want to find $\frac{du}{dv} = \frac{du / dx}{dv / dx}$. Clearly

$\frac{du}{dx} = 2 \sin x \cos x \quad \text{and} \quad \frac{dv}{dx} = e^{\cos x} (-\sin x) = -(\sin x) e^{\cos x}$

Thus $\frac{du}{dv} = \frac{2 \sin x \cos x}{-\sin x \cdot e^{\cos x}} = -\frac{2 \cos x}{e^{\cos x}}$