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

Answer

$y=x-1$ $y=\frac{1}{e}$
1556630486

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 $(e,1/e)$ $y^{\prime}=\frac{1}{e}\\ y=\frac{1}{e}$
Small 1556630486
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.