Answer
$y'=2xy+\frac{2}{\sqrt \pi} $
Work Step by Step
As we are given that $y=e^{x^{2}}erf(x)$
Since, the error function,$erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^{2}} dt$
Thus,
$y=e^{x^{2}}\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^{2}} dt$
Differentiating y with respect to x.
$y'=\frac{d}{dx}[e^{x^{2}}\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^{2}} dt]$
$y'=[e^{x^{2}}\frac{2}{\sqrt \pi} e^{-x^{2}} -0]+2xy$
Hence, $y'=2xy+\frac{2}{\sqrt \pi} $
