Answer
(a) $f'(x) = 4-sec^2~x$
$f''(x) = -2~sec^2~x~tan~x$
(b) We can see a sketch of the graphs below.
Work Step by Step
(a) $-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$
$f(x) = 4x-tan~x$
We can find $f'(x)$:
$f'(x) = 4-sec^2~x$
We can find $f''(x)$:
$f''(x) = (-2~sec~x)~(sec~x~tan~x)$
$f''(x) = -2~sec^2~x~tan~x$
(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.
$f''(x)$ 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.