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: 8

Answer

$\frac{dy}{dx} = (sec (\frac{1}{x} + 7x) \times tan (\frac{1}{x} + 7x))\times (\frac{1}{x^{2}} -7)$

Work Step by Step

$\frac{dy}{du} = \frac{d (-sec u)}{du} = -sec u\times tan u$ and, $\frac{du}{dx} = \frac{d (\frac{1}{x} + 7x)}{dx}$ $\frac{du}{dx} = \frac{-1}{x^{2}} +7$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = (-sec u\times tan u)(\frac{-1}{x^{2}} +7)$ $\frac{dy}{dx} = (-sec (\frac{1}{x} + 7x) \times tan (\frac{1}{x} + 7x))\times (\frac{-1}{x^{2}} +7)$ $\frac{dy}{dx} = (sec (\frac{1}{x} + 7x) \times tan (\frac{1}{x} + 7x))\times (\frac{1}{x^{2}} -7)$
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