Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 26


$$\frac{\cot \left(x\right)\csc \left(x\right)}{\left(\csc \left(x\right)-1\right)^2}$$

Work Step by Step

Given $$f(u)=\frac{u}{u-1}, \quad g(x)=\csc x$$ Since \begin{align*} f(g(x))&=\frac{g(x)}{g(x)-1}\\ &= \frac{\csc x }{\csc x-1} \end{align*} Then \begin{align*} f'(g(x))&= \frac{\frac{d}{dx}\left(\csc \left(x\right)\right)\left(\csc \left(x\right)-1\right)-\frac{d}{dx}\left(\csc \left(x\right)-1\right)\csc \left(x\right)}{\left(\csc \left(x\right)-1\right)^2}\\ &= \frac{\cot \left(x\right)\csc \left(x\right)}{\left(\csc \left(x\right)-1\right)^2} \end{align*}
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