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: 49


$$ y'= -4(3x^2-\sin x) (x^3+\cos x)^{-5}.$$

Work Step by Step

Since $ y=(x^3+\cos x)^{-4}$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ the derivative $ y'$ is given by: $$ y'=-4 (x^3+\cos x)^{-5}(x^3+\cos x)'\\=-4 (x^3+\cos x)^{-5}(3x^2-\sin x)\\=-4(3x^2-\sin x) (x^3+\cos x)^{-5}.$$ Here, we used the fact that $(\cos x)'=-\sin x $.
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