Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1023: 35


$\dfrac{du}{dp}=2xy^3(1+6p)+3x^2y^2(pe^p+e^p)+4z^3(p\cos p+\sin p)$

Work Step by Step

Given: $u=x^2y^3+z^4$ Apply chain rule. $\frac{du}{dp}=(2xy^3)\frac{dx}{dP}+(3x^2y^2)\frac{dy}{dP}+(4z^3)\frac{dz}{dp}$ Plug in the values of $x=p+3p^2,y=pe^p,z=p \sin p$ in $(2xy^3)\frac{dx}{dP}+(3x^2y^2)\frac{dy}{dP}+(4z^3)\frac{dz}{dP}$ $\dfrac{du}{dp}=2xy^3(1+6p)+3x^2y^2(pe^p+e^p)+4z^3(p\cos p+\sin p)$ Hence, $\dfrac{du}{dp}=2xy^3(1+6p)+3x^2y^2(pe^p+e^p)+4z^3(p\cos p+\sin p)$
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