University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 53



Work Step by Step

Given $y=(x+2)^{(x+2)}$ $lny=ln(x+2)^{(x+2)}$ $lny=(x+2)ln{(x+2)}$ On differentiating both sides: $\frac{1}{y}\frac{dy}{dx}=\ln{(x+2)}+\frac{(x+2)}{(x+2)}$ $\frac{dy}{dx}=y(\ln{(x+2)}+1)$ $\frac{dy}{dx}=(x+2)^{(x+2)}(\ln{(x+2)}+1)$
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