The differentiation of cos^2x w.r.to e^sin x is:

The correct answer is Option (2) → \(\displaystyle\frac{-2\sin x}{e^{\sin x}}\)

Let \(u=\cos^2x\) and \(v=e^{\sin x}\). To differentiate u with respect to v, divide the two derivatives with respect to x. By the chain rule:

    \( \displaystyle \frac{du}{dx} = 2\cos x\,(-\sin x) = -2\sin x\cos x \)

    \( \displaystyle \frac{dv}{dx} = e^{\sin x}\cos x \)

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

Cancelling \(\cos x\) (where \(\cos x \ne 0\)):

    \( \displaystyle \frac{du}{dv} = \frac{-2\sin x}{e^{\sin x}} \)