#### Answer

$f'(x)=4x^{3}-6x^{2}+2x$
The graphs of f'(x) correctly shows the increasing and decreasing nature of f(x), as well as all the vertices.

#### Work Step by Step

$f(x)=x^{4}-2x^{3}+x^{2}$
using the power rule, ($\frac{d}{dx}$[$x^{n}$]=n$x^{n-1}$) we can find that
$f'(x)=4x^{3}-6x^{2}+2x$
When you compare the graphs of f(x) and f'(x), they make sense. This is because from the left (negative), moving right, f(x) is decreasing until x=0. f'(x) is negative here and at x=0, crosses the x-axis. f(x) then increases from 0 to 1/2, and then decreases from 1/2 to 1. Through this, f'(x) is positive, zero, then negative, just as it should be. At x=1, f(x) has a vertex, meaning f(x) should have a zero, which it does. After x=1, f(x) increases and f'(x) is positive.