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: 26

Answer

$$f'(x)=-4(x^4-\sec(4x^2-2))^{-5}(4x^3-8x\sec(4x^2-2)\tan(4x^2-2))$$

Work Step by Step

$$f'(x)=\Big((x^4-\sec(4x^2-2))^{-4}\Big)'=-4(x^4-\sec(4x^2-2))^{-5}\cdot(x^4-\sec(4x^2-2))'= -4(x^4-\sec(4x^2-2))^{-5}(4x^3-\sec(4x^2-2)\tan(4x^2-2)\cdot(4x^2-2)')= -4(x^4-\sec(4x^2-2))^{-5}(4x^3-\sec(4x^2-2)\tan(4x^2-2)\cdot8x)= -4(x^4-\sec(4x^2-2))^{-5}(4x^3-8x\sec(4x^2-2)\tan(4x^2-2))$$
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