Question
If $f(x) = 3{e^{{x^2}}}$, then $f'(x) - 2xf(x) + {1 \over 3}f(0) - f'(0) = $
Differentiating w.r.t. $x,$ we get $f'(x) = 6x{e^{{x^2}}}$;
$\therefore f(0) = 3$ and $f'(0) = 0$
==>$f'(x) - 2xf(x) + \frac{1}{3}f(0) - f'(0)$
$ = 6x{e^{{x^2}}} - 6x{e^{{x^2}}} + \frac{1}{3}(3) - 0 = 1$ .
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