Calculus (3rd Edition)

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

Chapter 3 - Differentiation - Chapter Review Exercises - Page 164: 74



Work Step by Step

Recall that $(\tan x)'=\sec^2 x$. By differentiating the equation $ xy=\tan (x+y)$ with respect to $ x $, we get $$ y+x\frac{dy}{dx}=(1+\frac{dy}{dx})\sec^2(x+y)\\ \Longrightarrow \frac{dy}{dx}(x-\sec^2(x+y))=-y+\sec^2(x+y)$$ and hence $$\frac{dy}{dx}=\frac{-y+\sec^2(x+y)}{x-\sec^2(x+y))}.$$
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