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 - Page 158: 25

Answer

$$f'(x)=-3(x+\csc(x^3+3))^{-4}(1-3x^2\csc(x^3+3)\cot(x^3+3))$$

Work Step by Step

$$f'(x)=\Big((x+\csc(x^3+3))^{-3}\Big)'=-3(x+\csc(x^3+3))^{-4}\cdot(x+\csc(x^3+3))'= -3(x+\csc(x^3+3))^{-4}(1+(-\csc(x^3-3)\cot(x^3-3))\cdot(x^3+3)')= -3(x+\csc(x^3+3))^{-4}(1-\csc(x^3+3)\cot(x^3+3)\cdot3x^2)= -3(x+\csc(x^3+3))^{-4}(1-3x^2\csc(x^3+3)\cot(x^3+3))$$
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