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: 52


$$ y' =168x^3(5-2x^4)^6(9-(5-2x^4)^7)^2.$$

Work Step by Step

Since $ y=(9-(5-2x^4)^7)^3$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by $$ y'=3(9-(5-2x^4)^7)^2 (9-(5-2x^4)^7)' \\=3(9-(5-2x^4)^7)^2(-7(5-2x^4)^6(-8x^3))\\=168x^3(5-2x^4)^6(9-(5-2x^4)^7)^2.$$
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