Answer
$f'(x) = \frac{2x-1}{1+(x^2-x)^2}$
We can see a sketch of the graphs below.
Work Step by Step
$f(x) = arctan(x^2-x)$
We can find $f'(x)$:
$f'(x) = \frac{2x-1}{1+(x^2-x)^2}$
$f'(x)$ 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.