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: 23


$$ y'(x) = \frac{\sec ^{2} x-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-\sec ^{2} y} $$

Work Step by Step

Given $$\tan (x+y)=\tan x+\tan y$$ Differentiate with respect to $x$ \begin{align*} \frac{d}{dx} \tan (x+y)&=\frac{d}{dx}[\tan x+\tan y] \\ \sec^2 \left(x+ y\right)[1+y'(x) ]&=\sec^2 x+ \sec^2 y y'(x) \\ \left [\sec^2 \left(x+ y\right) -\sec^2 y \right] y'(x)& =\sec^2 x- \sec^2 \left(x+ y\right) \end{align*} Then $$ y'(x) = \frac{\sec ^{2} x-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-\sec ^{2} y} $$
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