Derivative of $\sin ^2 x$ with respect to $\mathrm{e}^{\cos x}$ is
Derivative of $\sin ^2 x$ with respect to $\mathrm{e}^{\cos x}$ is
- $2 \sin x \cos ^2 x \mathrm{e}^{\cos x}$
- $\frac{2 \cos x}{\mathrm{e}^{\cos x}}$
- $\frac{2 \sin x}{\mathrm{e}^{\cos x}}$
- $\frac{-2 \cos x}{e^{\cos x}}$
Solution
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
$\begin{aligned}
& \therefore \quad \frac{\mathrm{dv}}{\mathrm{~d} x}=-\mathrm{e}^{\cos x} \cdot \sin x \\
& \frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{2 \sin x \cdot \cos x}{-e^{\cos x} \cdot \sin x} \\
& =\frac{-2 \cos x}{\mathrm{e}^{\cos x}}
\end{aligned}$
Asked in: MHT CET 2024 (03 May Shift 2)
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