Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 965: 50


$\frac{\partial z}{\partial y}=\frac{-z - \frac{x}{y}}{y-2z}$ $\frac{\partial z}{\partial x}=\frac{\ln(y)}{y-2z}$

Work Step by Step

The function $yz+x\ln(y)=z^2$ is given. The partial derivatives of $z$ with respect to $x$ and $y$ need to be found. Partial derivative w.r.t $x$: $\frac{\partial}{\partial x} {(yz+x\ln(y))}= \frac{\partial}{\partial x} {z^2}$ $y \frac{\partial z}{\partial x} + \ln(y)=2z \frac{\partial z}{\partial x}$ Seperating the partial derivative: $\frac{\partial z}{\partial x}=\frac{\ln(y)}{y-2z}$ Partial derivative w.r.t $y$: $\frac{\partial}{\partial y} {(yz+x\ln(y))}= \frac{\partial}{\partial y} {z^2}$ $y \frac{\partial z}{\partial y} + z + \frac{x}{y}=2z \frac{\partial z}{\partial y}$ Seperating the partial derivative: $\frac{\partial z}{\partial y}=\frac{-z - \frac{x}{y}}{y-2z}$
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