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: 25


$s'(t) = \left(\dfrac{1215}{2}t^2\right)(3t^3-8)^{1/2}$

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 = 3t^3-8 $ $f(u) = 45u^{3/2}$ Derivate the function: $s'(u) = \dfrac{135}{2}u^{1/2}u'$ Now let's find u' $u' = 9t^2$ Then undo the substitution, simplify and get the answer: $s'(t) = \dfrac{135}{2}(9t^2)(3t^3-8)^{1/2}$ $s'(t) = \left(\dfrac{1215}{2}t^2\right)(3t^3-8)^{1/2}$
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