Answer
$$f'(x)= 5x^4 -6x^2 + 1$$
When comparing the graphs of $f(x)$ and $f'(x)$, $f'(x)$ makes sense because:
$f'(x)$ is positive when $f(x)$'s slope is positive
$f'(x)$ is negative when $f(x)$'s slope is negative
and $f'(x)$ is 0 when $f(x)$'s slope is 0.
Work Step by Step
$f(x)=x^5-2x^3+x-1$
Using the power rule, $\frac{d}{dx} \ x^n =nx^{n-1}$ we get:
$f'(x)= 5x^4 -6x^2 + 1$
See Image of Graph.
$f(x)$ is red
$f'(x)$ is blue
