Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 34



Work Step by Step

$y=x^{2} \ln x\\ y^{\prime}=\left(x^{2}\right)\left(\frac{1}{x}\right)+(\ln x)(2 x)\\ y^{\prime}=x+2 x \ln x\\ y^{\prime}=x(1+2 \ln x)$ Gradient at point $(1,0)$ $y^{\prime}=1(1+2 \ln 1)\\ y^{\prime}=1$ Tangent line at point $(1,0)$ $y-0=1(x-1)\\ y=x-1$
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