Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises - Page 437: 56

Answer

$y'=\frac{y(xlny-y)}{x(ylnx-x)}$

Work Step by Step

Given: $x^{y}=y^{x}$ Use logarithmic properties $ln(x^{y})=ylnx$. $ylnx=xlny$ Simplify the given function using both product rule and chain rule of differentiation. $\frac{y}{x}+lnx\frac{dy}{dx}=\frac{x}{y}\frac{dy}{dx}+lny$ or $\frac{y}{x}+lnxy'=\frac{x}{y}y'+lny$ Thus, $y'=\frac{lny-\frac{y}{x}}{lnx-\frac{x}{y}}$ Hence, $y'=\frac{y(xlny-y)}{x(ylnx-x)}$
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