If f(x) = 3e^{x^2},then f'(x) - 2xf(x) + frac | vedclass

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(B) Given the function $f(x) = 3e^{x^2}$.Differentiating with respect to $x$,we get $f'(x) = 3 \cdot e^{x^2} \cdot (2x) = 6xe^{x^2}$.Now,evaluate the

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$1$

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(B) Given the function $f(x) = 3e^{x^2}$.Differentiating with respect to $x$,we get $f'(x) = 3 \cdot e^{x^2} \cdot (2x) = 6xe^{x^2}$.Now,evaluate the function and its derivative at $x = 0$:$f(0) = 3e^{0^2} = 3(1) = 3$.$f'(0) = 6(0)e^{0^2} = 0$.Substitute these values into the expression $f'(x) - 2xf(x) + \frac{1}{3}f(0) - f'(0)$:$= 6xe^{x^2} - 2x(3e^{x^2}) + \frac{1}{3}(3) - 0$$= 6xe^{x^2} - 6xe^{x^2} + 1 - 0$$= 1$.

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

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