Answer
$$f'(g(x))= \sec^2 x-\csc^2x $$
Work Step by Step
Given $$f(u)=u+u^{-1}, \quad g(x)=\tan x$$
Since
\begin{align*}
f(g(x))&=\tan x+ (\tan x)^{-1}\\
&= \tan x+ \cot x
\end{align*}
Then
$$f'(g(x))= \sec^2 x-\csc^2x $$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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: 25
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