Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises: 66

Answer

$\frac{dy}{dx}=(\frac{x-(1+x)ln(1+x)}{x^2(1+x)})(1+x)^{\frac{1}{x}}$

Work Step by Step

$y=(1+x)^{\frac{1}{x}}$ $lny=ln(1+x)^{\frac{!}{x}}$ $lny=\frac{1}{x}ln(1+x)$ $\frac{dy}{y}=(-\frac{1}{x^2}ln(1+x)+\frac{1}{x(1+x)})dx$ $\frac{dy}{dx}=(\frac{-(1+x)ln(1+x)+x}{x^2(1+x)})y$ $=(\frac{x-(1+x)ln(1+x)}{x^2(1+x)})(1+x)^{\frac{1}{x}}$$
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