Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 421: 100


$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} $
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