Find the derivative: frac{d}{dx}(x e^{x^2}) = | vedclass

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(D) To find the derivative of $f(x) = x e^{x^2}$,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.Let $u(x) = x$ and $v(x) =

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None of these

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(D) To find the derivative of $f(x) = x e^{x^2}$,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.Let $u(x) = x$ and $v(x) = e^{x^2}$.Then $u'(x) = 1$ and $v'(x) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x = 2x e^{x^2}$.Applying the product rule:$\frac{d}{dx}(x e^{x^2}) = (1)(e^{x^2}) + (x)(2x e^{x^2})$$= e^{x^2} + 2x^2 e^{x^2}$$= (1 + 2x^2)e^{x^2}$.Comparing this with the given options,none of the options $A, B, C$ match the result $(1 + 2x^2)e^{x^2}$.Therefore,the correct choice is $D$.

Find the derivative: $\frac{d}{dx}(x e^{x^2}) = $

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