7. The four real roots of the quartic polynomial $f(x)$ form an arithmetic sequence with a common difference of 2. Then, the difference between the largest and smallest roots of $f^{\prime}(x)$ is
7. $2 \sqrt{5}$.

Let the four real roots of $f(x)$ be
$$
\begin{array}{l} 
a-3, a-1, a+1, a+3. \\
\text { Then } f(x)= k(x-a-3)(x-a-1) . \\
(x-a+1)(x-a+3)(k \neq 0) .
\end{array}
$$

Let $x-a=t$. Then
$$
\begin{array}{l}
f(x)=k(t-3)(t-1)(t+1)(t+3) \\
=k\left(t^{4}-10 t^{2}+9\right)=g(t) .
\end{array}
$$

Thus, $g^{\prime}(t)=k\left(4 t^{3}-20 t\right)=4 k\left(t^{3}-5 t\right)$.
The roots of $g^{\prime}(t)=0$ are $\pm \sqrt{5}, 0$, so the roots of $f^{\prime}(x)$ are $a \pm \sqrt{5}, a$.
Therefore, the difference between the largest and smallest roots of $f^{\prime}(x)$ is $2 \sqrt{5}$.