Question

Differentiate the function \(\sin^2x\) with respect to the function \(e^{\cos x}\).

Hint:

Use \(\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}\), or rewrite \(u\) directly in terms of \(t=\cos x\).

Answer 1

Step 1: Writing u directly as a function of v:
Let \(t=\cos x\), so \(v=e^{t}\) and \(u=\sin^2x=1-\cos^2x=1-t^2\).

Step 2: Differentiating u w.r.t. t, and v w.r.t. t:
\(\dfrac{du}{dt}=-2t\), and \(\dfrac{dv}{dt}=e^{t}\).

Step 3: Chain-dividing:
\(\dfrac{du}{dv}=\dfrac{du/dt}{dv/dt}=\dfrac{-2t}{e^{t}}=-2t\,e^{-t}\). Substituting back \(t=\cos x\): \(\dfrac{du}{dv}=-2\cos x\,e^{-\cos x}\).

Final Answer:
Same result: \(\boxed{-2\cos x\,e^{-\cos x}}\).