Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 914: 46

Answer

$$\frac{{\partial z}}{{\partial x}} = y$$

Work Step by Step

$$\eqalign{ & xy - z = 1 \cr & {\text{Solve for }}z \cr & z = xy - 1 \cr & {\text{Find }}\frac{{\partial z}}{{\partial x}},{\text{ differentiate both sides}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {xy - 1} \right] \cr & \frac{{\partial z}}{{\partial x}} = y \cr} $$
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