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: 23


$k'(x) = (288x)(12x^2+5)^{-7}$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make an «u» substitution to make it easier $u = 12x^2+5$ $k(u) = -2u^{-6}$ Derivate the function: $k'(u) = 12u^{-7}7u'$ Now let's find u' $u' = 24x$ Then undo the substitution, simplify and get the answer: $k'(x) = 12(24x)(12x^2+5)^{-7}$ $k'(x) = (288x)(12x^2+5)^{-7}$
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