University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 15


$$\frac{dy}{dx}=\sec(\tan x)\tan(\tan x)\sec^2x$$

Work Step by Step

$$y=f(x)=\sec(\tan x)$$ Let $u=\tan x$, then $y=f(u)=\sec u$ We have $$\frac{dy}{du}=\frac{d}{du}(\sec u)=\sec u\tan u$$ $$\frac{du}{dx}=\frac{d}{dx}(\tan x)=\sec^2x$$ According to the Chain Rule: $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\sec u\tan u\times\sec^2 x$$ Replace $u$ back with $\tan x$: $$\frac{dy}{dx}=\sec(\tan x)\tan(\tan x)\sec^2x$$
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