Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 27

Answer

$$\frac{\partial}{\partial p}\tan^{-1}(pq^2)=\frac{q^2}{1+p^2q^4}.$$ $$\frac{\partial}{\partial q}\tan^{-1}(pq^2)=\frac{2pq}{1+p^2q^4}.$$

Work Step by Step

Regard $q$ as constant and differentiate with respect to $p$: $$\frac{\partial}{\partial p}\tan^{-1}(pq^2)=\frac{1}{1+(pq^2)^2}\frac{\partial}{\partial p}(pq^2)=\frac{1}{1+p^2q^4}q^2=\frac{q^2}{1+p^2q^4}.$$ Regard $p$ as constant and differentiate witth respect to $q$: $$\frac{\partial}{\partial q}\tan^{-1}(pq^2)=\frac{1}{1+(pq^2)^2}\frac{\partial}{\partial q}(pq^2)=\frac{1}{1+p^2q^4}2pq=\frac{2pq}{1+p^2q^4}.$$
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