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: 15

Answer

$g_u(0,0)=7,g_v(0,0)=2$

Work Step by Step

Notice that $g(0,0)=f(e^0+\sin0,e^0+\cos0)=f(1,2)$. We will use the Chain Rule to find $g_u(0,0)$ and $g_v(0,0)$: $$g_u(0,0)=f_x(1,2)\left.\frac{\partial}{\partial u}(e^u+\sin v)\right|_{(u,v)=(0,0)}+f_y(1,2)\left.\frac{\partial}{\partial u}(e^u+\cos v)\right|_{(u,v)=(0,0)}= 2\cdot e^0+5\cdot e^0=7$$ $$g_v(0,0)=f_x(1,2)\left.\frac{\partial}{\partial v}(e^u+\sin v)\right|_{(u,v)=(0,0)}+f_y(1,2)\left.\frac{\partial}{\partial v}(e^u+\cos v)\right|_{(u,v)=(0,0)}= 2\cdot\cos0+5\cdot(-\sin 0)=2$$
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