University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 90


$$y'=x^{x+1}\Big(\ln x+\frac{x+1}{x}\Big)$$

Work Step by Step

$$y=x^{x+1}$$ We note that $x^{x+1}=e^{\ln(x^{x+1})}=e^{(x+1)\ln x}$ Therefore, $$y=e^{(x+1)\ln x}$$ The derivative of $y$ is: $$y'=(e^{(x+1)\ln x})'=e^{(x+1)\ln x}((x+1)\ln x)'$$ $$y'=e^{(x+1)\ln x}(\ln x+(x+1)\times\frac{1}{x})$$ $$y'=e^{(x+1)\ln x}\Big(\ln x+\frac{x+1}{x}\Big)$$ $$y'=x^{x+1}\Big(\ln x+\frac{x+1}{x}\Big)$$
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