Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 27


$g'(t) = -\dfrac{63}{2}t^2(7t^3-1)^{-1/2} $

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make a «u» substitution to make it easier $u = 7t^3-1 $ $g(u) = -3u^{1/2}$ Derivate the function: $g'(u) = -\dfrac{3}{2}u^{-1/2}u'$ Now let's find u' $u' = 21t^2$ Then undo the substitution, simplify and get the answer: $g'(t) = -\dfrac{3}{2}(21t^2)(7t^3-1)^{-1/2} $ $g'(t) = -\dfrac{63}{2}t^2(7t^3-1)^{-1/2} $
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