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: 63


$$ y'=-180x^3(1+\cot^5(x^4+1))^8\cot^4(x^4+1)\csc^2(x^4+1).$$

Work Step by Step

Since $ y=(1+\cot^5(x^4+1))^9$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\cot x)'=-\csc^2 x $, the derivative $ y'$ is given by $$ y'=9(1+\cot^5(x^4+1))^8(1+\cot^5(x^4+1))' =9(1+\cot^5(x^4+1))^8(5\cot^4(x^4+1))(\cot^4(x^4+1))' \\ \\=9(1+\cot^5(x^4+1))^8(5\cot^4(x^4+1)(-4x^3\csc^2(x^4+1)))\\ =-180x^3(1+\cot^5(x^4+1))^8\cot^4(x^4+1)\csc^2(x^4+1).$$
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