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: 65


$$ y' =-36x^2\cot (1-x^3)\csc^2(1-x^3)(1-\csc^2(1-x^3))^5.$$

Work Step by Step

Since $ y=(1-\csc^2(1-x^3))^6$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\csc x)'=-\csc x\cot x $, the derivative $ y'$ is given by $$ y'=6(1-\csc^2(1-x^3))^5(1-\csc^2(1-x^3))'\\ =6(1-\csc^2(1-x^3))^5(-2\csc(1-x^3).\\(-\csc (1-x^3)\cot (1-x^3)(-3x^2)))\\ =-36x^2\cot (1-x^3)\csc^2(1-x^3)(1-\csc^2(1-x^3))^5.$$
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