Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 781: 24


$$ S_w =\frac{z}{1+(wz)^2}, \quad S_z=\frac{w}{1+(wz)^2}. $$

Work Step by Step

Recall that $(\tan^{-1} x)'=\dfrac{1}{1+x^2}$ Since $ S=\tan^{-1}(wz)$, then using the chain rule, we have $$ S_w =\frac{z}{1+(wz)^2}, \quad S_z=\frac{w}{1+(wz)^2}. $$
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