Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 212: 89

Answer

$$\frac{d}{{dx}}{\left( {1 + \frac{1}{x}} \right)^x} = {\left( {1 + \frac{1}{x}} \right)^x}\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right)$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}{\left( {1 + \frac{1}{x}} \right)^x} \cr & or \cr & \frac{d}{{dx}}{\left( {\frac{{x + 1}}{x}} \right)^x} \cr & {\text{let }}y = {\left( {\frac{{x + 1}}{x}} \right)^x} \cr & {\text{taking the natural logarithm of both sides of the equation}} \cr & \,\,\,\ln y = \ln {\left( {\frac{{x + 1}}{x}} \right)^x} \cr & {\text{using logarithmic properties}} \cr & \,\,\,\ln y = x\ln \left( {\frac{{x + 1}}{x}} \right) \cr & \,\,\,\ln y = x\ln \left( {x + 1} \right) - x\ln x \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {x\ln \left( {x + 1} \right)} \right] - \frac{d}{{dx}}\left[ {x\ln \left( x \right)} \right] \cr & {\text{compute derivatives}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {\ln \left( {x + 1} \right)} \right] + \ln \left( {x + 1} \right)\frac{d}{{dx}}\left[ x \right] - x\frac{d}{{dx}}\left[ {\ln \left( x \right)} \right] - \ln \left( x \right)\frac{d}{{dx}}\left[ x \right] \cr & \frac{1}{y}\frac{{dy}}{{dx}} = x\left( {\frac{1}{{x + 1}}} \right) + \ln \left( {x + 1} \right) - x\left( {\frac{1}{x}} \right) - \ln \left( x \right) \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{{x + 1}} + \ln \left( {x + 1} \right) - 1 - \ln \left( x \right) \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{{x + 1}} + \ln \left( {\frac{{x + 1}}{x}} \right) - 1 + \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}} \cr & {\text{solving for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = y\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right) \cr & {\text{replace }}y{\text{ with }}{\left( {1 + \frac{1}{x}} \right)^x} \cr & \frac{d}{{dx}}{\left( {1 + \frac{1}{x}} \right)^x} = {\left( {1 + \frac{1}{x}} \right)^x}\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right) \cr} $$
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.