Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises: 1

Answer

$$\frac{dz}{dt}=2t(4t^6-9t^4-2t^2+3)$$

Work Step by Step

$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=\frac{\partial}{\partial x}(xy^3-x^2y)\frac{d}{dt}(t^2+1)+\frac{\partial }{\partial y}(xy^3-x^2y)\frac{d}{dt}(t^2-1)= (y^3-2xy)\cdot2t+(3xy^2-x^2)\cdot2t=2t(y^3-2xy+3xy^2-x^2)$$ Expressing this in terms of $t$ we have: $$\frac{dz}{dt}=2t(y^3-2xy+3xy^2-x^2)=2t((t^2-1)^3-2(t^2+1)(t^2-1)+3(t^2+1)(t^2-1)^2-(t^2+1)^2)= 2t(t^6-3t^4+3t^2-1-2t^4+2+3t^6-3t^4-3t^2+3-t^4-2t^2-1)= 2t(4t^6-9t^4-2t^2+3)$$
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