Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises: 5

Answer

$$\frac{dw}{dt}=e^{\frac{1-t}{1+2t}}\frac{t(2+5t+8t^2)}{(1+2t)^2}$$

Work Step by Step

$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}=\frac{\partial}{\partial x}(xe^{y/z})\frac{d}{dt}(t^2)+\frac{\partial}{\partial y}(xe^{y/z})\frac{d}{dt}(1-t)+\frac{\partial}{\partial z}(xe^{y/z})\frac{d}{dt}(1+2t)= e^{y/z}\cdot2t+xe^{y/z}\frac{\partial}{\partial y}(y/z)\cdot(-1)+xe^{y/x}\frac{\partial}{\partial z}(y/z)\cdot2= 2te^{y/z}-xe^{y/z}\frac{1}{z}+2xe^{y/z}\cdot\Big(-\frac{y}{z^2}\Big)= e^{y/z}\Big(2t-\frac{x}{z}-\frac{2xy}{z^2}\Big)$$ Now we have to express solution in terms of $t$: $$\frac{dw}{dt}=e^{y/z}\Big(2t-\frac{x}{z}-\frac{2xy}{z^2}\Big)= e^{\frac{1-t}{1+2t}}\Big(2t-\frac{t^2}{1+2t}-\frac{2t^2(1-t)}{(1+2t)^2}\Big)= e^{\frac{1-t}{1+2t}}\frac{2t(1+2t)^2-t^2(1+2t)-2t^2+2t^3}{(1+2t)^2}= e^{\frac{1-t}{1+2t}}\frac{2t+8t^2+8t^3-t^2-2t^3-2t^2+2t^3}{(1+2t)^2}= e^{\frac{1-t}{1+2t}}\frac{t(2+5t+8t^2)}{(1+2t)^2}$$
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