Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 2


$f'(u) = 24x^2(3x^3+5)^3$

Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$. Original expression: $y = (2x^3 + 5)^4$ $u = g(x) = 2x^3 + 5$ $y = f(u) = x^4$ Apply the chain rule to find the derivative: $f'(g(x)) \times g'(x) $ = > $f'(u) \times u'$ $\frac{dy}{du}(u)^4 \times\frac{du}{dx}(2x^3+5) $ Apply the power rule $4(2x^3 + 5)^3\times6x^2$ $24x^2(2x^3+5)^3$
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