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: 16


$\frac{dy}{dx} = -cosec^{2} (\pi - \frac{1}{x}) \times \frac{1}{x^{2}}$

Work Step by Step

$y =cot u$ where, $u = (\pi - \frac{1}{x})$ Now, $\frac{dy}{du} = -cosec^{2} u$ and, $\frac{du}{dx} =\frac{1}{x^{2}}$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = -cosec^{2} u \times \frac{1}{x^{2}}$ $\frac{dy}{dx} = -cosec^{2} (\pi - \frac{1}{x}) \times \frac{1}{x^{2}}$
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