If f(x)=3 e^{x^2} then f^{prime}(x)-2 x f(x)+ | vedclass

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(B) Given $f(x) = 3 e^{x^2}$.First,find the derivative $f^{\prime}(x)$ using the chain rule:$f^{\prime}(x) = 3 \cdot e^{x^2} \cdot \frac{d}{dx}(x^2) =

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(B) Given $f(x) = 3 e^{x^2}$.First,find the derivative $f^{\prime}(x)$ using the chain rule:$f^{\prime}(x) = 3 \cdot e^{x^2} \cdot \frac{d}{dx}(x^2) = 3 e^{x^2} \cdot 2x = 6x e^{x^2}$.Now,calculate the terms:$2x f(x) = 2x(3 e^{x^2}) = 6x e^{x^2}$.$f(0) = 3 e^{0^2} = 3(1) = 3$.$f^{\prime}(0) = 6(0) e^{0^2} = 0$.Substitute these into the expression $f^{\prime}(x) - 2x f(x) + \frac{1}{3} f(0) - f^{\prime}(0)$:$= 6x e^{x^2} - 6x e^{x^2} + \frac{1}{3}(3) - 0$$= 0 + 1 - 0 = 1$.

If $f(x)=3 e^{x^2}$ then $f^{\prime}(x)-2 x f(x)+\frac{1}{3} f(0)-f^{\prime}(0)=$

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