Question

Prove the following identity:
$f^{\prime }\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^{\prime }\left(0\right)=1$, if $f\left(x\right)=3e^{x3}.$

Answer 1

$f\left(x\right)=3e^{x^2}\Rightarrow f\left(0\right)=3$

$f^{\prime }\left(x\right)=6x{e}^{x^2}\Rightarrow f^{\prime }\left(0\right)=0$

So $=f^{\prime }\left(x\right)-2xf\left(x\right)+\frac{1}{3}f\left(0\right)-f^{\prime }\left(0\right)$

$=6x{e}^{x^2}-2x\left(3e^{x^2}\right)+\frac{1}{3}\left(3\right)-0$

$\Rightarrow 6x{e}^{x^2}-6x{e}^{x^2}+1$

$\Rightarrow 1$.