Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 817: 9

Answer

w = xy + yz + xz, x = u + y, y = u - y, z = uy; (u, y) = (1/2, 1) $\frac{dw}{du}$=$\frac{dw}{dx}$.$\frac{dx}{du}$+$\frac{dw}{dy}$.$\frac{dy}{du}$+$\frac{dw}{dz}$.$\frac{dz}{du}$ $\frac{dw}{du}$=(y+z).1+(x+z).1+(y+x).(v) $\frac{dw}{du}$=y+z+x+z+v(y+x) =2z+x+y+v(y+x) =2uv+u+v+u-v+v(u-v+u+v) =2uv+2u+2uv =2$\times$$\frac{1}{2}$+2$\times$$\frac{1}{2}$+2$\times$$\frac{1}{2}\times1$ =1+1+1 =3 $\frac{dw}{dv}$=$\frac{dw}{dx}$.$\frac{dx}{dv}$+$\frac{dw}{dy}$.$\frac{dy}{dv}$+$\frac{dw}{dz}$.$\frac{dz}{dv}$ $\frac{dw}{du}$=(y+z).1+(x+z)(-1)+(y+x).(u) =y+z-x-z=(x+y)u =u-v-u-v+(u+v+u-v).u =-2v+2$u^{2}$ =-2+2$\times$$\frac{1}{2}^{2}$ =$\frac{-3}{2}$
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