Answer
$$f'(x)=\sec x\tan x-1 $$
Work Step by Step
Given
$$ f(x)= \sec x-x$$
Since
\begin{aligned}
f'(x)&= \sec x\tan x-1
\end{aligned}
We can note that $f(x) $ is decreasing when $f'(x)<0$ and $f(x) $ is increasing when $f'(x)>0 $
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 150: 27
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