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: 44


The solution is $$f_z(e,1,0)=1.$$

Work Step by Step

Lets first find the derivative: $$f_z=\Big(x^{yz}\Big)'_z=\Big(\big(x^y\big)^z\Big)'_z=(x^y)^z\ln x^y=x^{yz}y\ln x,$$ where we used the fact that $x^{yz}=(x^y)^z$ and also the logarithmic rule $\ln x^y=y\ln x$. Now let us calculate by direct substitution $x=e$, $y=1$ and $z=0$: $$f_z(e,1,0)=e^{1\cdot0}=e^0=1.$$
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