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: 36


$y=x-1$ $y=\frac{1}{e}$

Work Step by Step

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