Answer
$$ f'(x)=\frac{8x}{4x^2+1}.$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Since $ f(x)= \ln(4x^2+1)$, then the derivative is given by
$$ f'(x)=\frac{1}{4x^2+1} (4x^2+1)'=\frac{8x}{4x^2+1}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 14
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