Derivative of sin ^2 x with respect to e^{cos | vedclass

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(D) Let $u = \sin ^2 x$ and $v = e^{\cos x}$.First,differentiate $u$ with respect to $x$:$\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cdot \co

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $u = \sin ^2 x$ and $v = e^{\cos x}$.First,differentiate $u$ with respect to $x$:$\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cdot \cos x$.Next,differentiate $v$ with respect to $x$:$\frac{dv}{dx} = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} \cdot (-\sin x) = -\sin x \cdot e^{\cos x}$.Now,find the derivative of $u$ with respect to $v$ using the chain rule:$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2 \sin x \cdot \cos x}{-\sin x \cdot e^{\cos x}}$.Canceling $\sin x$ from the numerator and denominator,we get:$\frac{du}{dv} = \frac{2 \cos x}{-e^{\cos x}} = \frac{-2 \cos x}{e^{\cos x}}$.

Derivative of $\sin ^2 x$ with respect to $e^{\cos x}$ is

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