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MHT CET · Maths · Differentiation

Derivative of \(\sin ^2 x\) with respect to \(\mathrm{e}^{\cos x}\) is

  1. A \(2 \sin x \cos ^2 x \mathrm{e}^{\cos x}\)
  2. B \(\frac{2 \cos x}{\mathrm{e}^{\cos x}}\)
  3. C \(\frac{2 \sin x}{\mathrm{e}^{\cos x}}\)
  4. D \(\frac{-2 \cos x}{e^{\cos x}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{-2 \cos x}{e^{\cos x}}\)

Step-by-step Solution

Detailed explanation

Let \(\mathrm{u}=\sin ^2 x, \mathrm{v}=\mathrm{e}^{\cos x}\)
\(\mathrm{u}=\sin ^2 x\)
Differentiating w.r.t. \(x\), we get
\(\frac{\mathrm{du}}{\mathrm{~d} x}=2 \sin x \cdot \cos x\)
Consider, \(\mathrm{v}=\mathrm{e}^{\cos x}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{dv}}{\mathrm{~d} x}=-\mathrm{e}^{\cos x} \cdot \sin x \\
& \frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{2 \sin x \cdot \cos x}{-e^{\cos x} \cdot \sin x} \\
& =\frac{-2 \cos x}{\mathrm{e}^{\cos x}}
\end{aligned}\)