Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 34

Answer

$$\frac{dy}{dx}=\frac{-2x\csc x^2\cot x^2(1-\cot x^2)-(1+\csc x^2)2x\csc ^2x^2}{(1-\cot x^2)^2}$$

Work Step by Step

$$\frac{dy}{dx}=\Big(\frac{1+\csc x^2}{1-\cot x^2}\Big)'= \frac{(1+\csc x^2)'(1-\cot x^2)-(1+\csc x^2)(1-\cot x^2)'}{(1-\cot x^2)^2}= \frac{(-\csc x^2\cot x^2\cdot(x^2)')(1-\cot x^2)-(1+\csc x^2)(\csc^2x^2\cdot(x^2)')}{(1-\cot x^2)^2}= \frac{-2x\csc x^2\cot x^2(1-\cot x^2)-(1+\csc x^2)2x\csc ^2x^2}{(1-\cot x^2)^2}$$
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