Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 975: 28

Answer

$dT=\dfrac{-v(vw)}{(1+uvw)^2}du+\dfrac{1+uvw-v(uw)}{(1+uvw)^2} dv+\dfrac{-v(uv)}{(1+uvw)^2} dw=\dfrac{1}{(1+uvw)^2}(-v^2w du+dv-uv^2dw)$

Work Step by Step

Here, we have $T_u=\dfrac{(0v) (vw)}{(1+uvw)^2} ; \\T_v=\dfrac{1+uvw-v(uw)}{(1+uvw)^2} ; \\T_w=\dfrac{(-v) (uv)}{(1+uvw)^2}$ $dT=\dfrac{\partial T}{\partial u} du +\dfrac{\partial T}{\partial v} dv+ \dfrac{\partial T}{\partial w} dw \\=\dfrac{-v(vw)}{(1+uvw)^2}du+\dfrac{1+uvw-v(uw)}{(1+uvw)^2} dv+\dfrac{-v(uv)}{(1+uvw)^2} dw \\=\dfrac{1}{(1+uvw)^2}(-v^2w du+dv-uv^2dw)$
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