## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 52

#### Answer

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

#### 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 $\frac{y}{x}+lnxy'=\frac{x}{y}y'+lny$ Hence, $y'=\frac{lny-\frac{y}{x}}{lnx-\frac{x}{y}}$ or $y'=\frac{y(xlny-y)}{x(ylnx-x)}$

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