Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 38


$$ y'=\frac{2\ln x}{x}e^{(\ln x)^2}.$$

Work Step by Step

Recall that $(e^x)'=e^x$ Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(x^n)'=nx^{n-1}$ Since $ y=e^{(\ln x)^2}$, then by the chain rule, we have $$ y'=e^{(\ln x)^2}(2\ln x)(1/x)=\frac{2\ln x}{x}e^{(\ln x)^2}.$$
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