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: 26

Answer

$s'(t) =(144t^3)(2t^4+5)^{1/2}$

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 = 2t^4+5 $ $f(u) =12u^{3/2} $ Derivate the function: $s'(u) = 18u^{1/2}u'$ Now let's find u' $u' = 8t^3$ Then undo the substitution, simplify and get the answer: $s'(t) = 18(8t^3)(2t^4+5)^{1/2}$ $s'(t) =(144t^3)(2t^4+5)^{1/2}$
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