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: 36


$$ y' =x(\ln x^2 + 1)$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(x^n)'=nx^{n-1}$ Since $ y=x^2 \ln x $, then using the product rule we have $$ y'=2x \ln x+ x^2 \frac{1}{x}=2x \ln x +x=x(\ln x^2 + 1)$$
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