Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 111


$$ y^{\prime}=(x+1)^{x}\left[\frac{x}{x+1}+\ln (x+1)\right]$$

Work Step by Step

Given $$ y=(x+1)^{x} $$ So, we have: \begin{aligned} &y=(x+1)^{x} \\ &\Rightarrow \ln y=\ln (x+1)^{x} \\ &\Rightarrow \ln y=x \ln (x+1) \\ &\text{ differentiate with respect to } \ x,\\ &\Rightarrow \frac{y^{\prime}}{y}=\ln (x+1)+x \cdot \frac{1}{(x+1)} \\& \Rightarrow y^{\prime}=y\left[\ln (x+1)+ \frac{x}{(x+1)} \right] \\& \Rightarrow y^{\prime}=(x+1)^{x}\left[\frac{x}{x+1}+\ln (x+1)\right] \end{aligned}
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