Derivative of Error Function

Theorem

 * $\displaystyle \frac \d {\d x} \paren {\map \erf x} = \frac 2 {\sqrt \pi} e^{-x^2}$

where $\erf$ denotes the error function.

Proof
We have, by the definition of the error function:


 * $\displaystyle \map \erf x = \frac 2 {\sqrt \pi} \int_0^x e^{-t^2} \rd t$

By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have:


 * $\displaystyle \frac \d {\d x} \paren {\map \erf x} = \frac 2 {\sqrt \pi} e^{-x^2}$