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: 16


$f'(x) = -\dfrac{4(3x^2+3)}{3(x^3+3x+9)^{-7/3}}$

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 $u = x^3+3x+9$ $f(u) = u^{-4/3}$ Derivate the function: $f'(u) = -\dfrac{4}{3}u^{-7/3}u'$ Now let's find u' $u' = 3x^2+3$ Then undo the substitution, simplify and get the answer: $f'(x) = -\dfrac{4(3x^2+3)}{3(x^3+3x+9)^{-7/3}}$
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