Differentiate sin ^{2} x with respect to e^{c | vedclass

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Question

(A) Let $u(x) = \sin ^{2} x$ and $v(x) = e^{\cos x}$.We need to find the derivative of $u$ with respect to $v$,which is given by $\frac{du}{dv} = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $u(x) = \sin ^{2} x$ and $v(x) = e^{\cos x}$.We need to find the derivative of $u$ with respect to $v$,which is given by $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.First,differentiate $u(x)$ with respect to $x$ using the chain rule:$\frac{du}{dx} = \frac{d}{dx}(\sin ^{2} x) = 2 \sin x \cos x$.Next,differentiate $v(x)$ with respect to $x$ using the chain rule:$\frac{dv}{dx} = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} \cdot \frac{d}{dx}(\cos x) = e^{\cos x} \cdot (-\sin x) = -\sin x e^{\cos x}$.Now,substitute these into the formula for $\frac{du}{dv}$:$\frac{du}{dv} = \frac{2 \sin x \cos x}{-\sin x e^{\cos x}}$.Assuming $\sin x 
eq 0$,we can cancel $\sin x$ from the numerator and denominator:$\frac{du}{dv} = \frac{2 \cos x}{-e^{\cos x}} = -2 \cos x e^{-\cos x}$.

Differentiate $\sin ^{2} x$ with respect to $e^{\cos x}$.

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