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: 72


$$ y'= -2kt^3(kt^4+b)^{-3/2}.$$

Work Step by Step

Recall that $(x^n)'=nx^{n-1}$ Since $ y=\frac{1}{ \sqrt{kt^4+b}}$, we rewrite $ y $ as follows $$ y=(kt^4+b)^{-1/2}.$$ Now, by using the chain rule, the derivative $ y'$ is given by $$ y'=-\frac{1}{2}(kt^4+b)^{-3/2}(4kt^3)=-2kt^3(kt^4+b)^{-3/2}.$$
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