Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 6


$$\frac{d}{dx}\tan(xy)= (y+x\frac{dy}{dx})\sec^2 (xy).$$

Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\tan x)'=\sec^2 x$. Using the chain rule, we have $$\frac{d}{dx}\tan(xy)= (y+x\frac{dy}{dx})\sec^2 (xy).$$
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