## Calculus 8th Edition

$y'=2xy+\frac{2}{\sqrt \pi}$
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, $y'=2xy+\frac{2}{\sqrt \pi}$