Answer
(a) $f'(x) = sec~x~tan~x-1$
(b) We can see a sketch of the graphs below.
Work Step by Step
(a) $f(x) = sec~x-x$
$f'(x) = sec~x~tan~x-1$
(b) We can see a sketch of the graphs below.
$f'(x)$ seems reasonable since $f'(x)$ has negative values when the slope of $f(x)$ has a negative slope, $f'(x)$ is 0 when the slope of $f(x)$ is 0, and $f'(x)$ has positive values when the slope of $f(x)$ has a positive slope.
