Answer
$f'(x) = \frac{x^2}{8}$
Work Step by Step
$\frac{d}{dx}[f(2x)] = x^2$
$f(2x) = \int x^2~dx$
$f(2x) = \frac{x^3}{3}+C$
$f(x) = \frac{(\frac{x}{2})^3}{3}+C$
$f(x) = \frac{x^3}{24}+C$
$f'(x) = \frac{x^2}{8}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 3 - Review - Exercises - Page 269: 111
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