Find the derivative: frac{d}{dx} { e^{-ax^2} | vedclass

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(C) To find the derivative of the product of two functions,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.Let $u(x) = e^{-a

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$e^{-ax^2}(\cot x - 2ax \log \sin x)$

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(C) To find the derivative of the product of two functions,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.Let $u(x) = e^{-ax^2}$ and $v(x) = \log(\sin x)$.First,find the derivatives of $u(x)$ and $v(x)$:$u'(x) = e^{-ax^2} \cdot \frac{d}{dx}(-ax^2) = e^{-ax^2} \cdot (-2ax) = -2ax e^{-ax^2}$.$v'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.Now,apply the product rule:$\frac{d}{dx} \{ e^{-ax^2} \log(\sin x) \} = (-2ax e^{-ax^2}) \cdot \log(\sin x) + e^{-ax^2} \cdot \cot x$.Factor out $e^{-ax^2}$:$= e^{-ax^2} [\cot x - 2ax \log(\sin x)]$.Thus,the correct option is $C$.

Find the derivative: $\frac{d}{dx} \{ e^{-ax^2} \log(\sin x) \}$

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