Derivative of Error Function
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Theorem
- $\map {\dfrac \d {\d x} } {\map \erf x} = \dfrac 2 {\sqrt \pi} e^{-x^2}$
where $\erf$ denotes the error function.
Proof
We have, by the definition of the error function:
- $\ds \map \erf x = \frac 2 {\sqrt \pi} \int_0^x e^{-t^2} \rd t$
By Fundamental Theorem of Calculus (First Part): Corollary, we therefore have:
- $\map {\dfrac \d {\d x} } {\map \erf x} = \dfrac 2 {\sqrt \pi} e^{-x^2}$
$\blacksquare$