Answer
Let $f( x)$ is a differentiable function then we have
$$
x f(x)]^{\prime}=x f^{\prime}(x)+f(x)'
$$
$$
\begin{aligned}
x f(x)]^{\prime \prime} &=x f^{\prime \prime}(x)+f^{\prime}(x)+f^{\prime}(x) \\
&=x f^{\prime \prime}(x)+2 f^{\prime}(x),
\end{aligned}
$$
and
$$
\begin{aligned}
x f(x)]^{\prime \prime \prime} &=x f^{\prime \prime \prime}(x)+f^{\prime \prime}(x)+2 f^{\prime \prime}(x) \\
& =x f^{\prime \prime \prime}(x)+3 f^{\prime \prime}(x),
\end{aligned}
$$
and so on. In general, we get:
$$
[x f(x)]^{(n)}=x f^{(n)}(x)+n f^{(n-1)}(x).
$$
Work Step by Step
Let $f( x)$ is a differentiable function then we have
$$
x f(x)]^{\prime}=x f^{\prime}(x)+f(x)'
$$
$$
\begin{aligned}
x f(x)]^{\prime \prime} &=x f^{\prime \prime}(x)+f^{\prime}(x)+f^{\prime}(x) \\
&=x f^{\prime \prime}(x)+2 f^{\prime}(x),
\end{aligned}
$$
and
$$
\begin{aligned}
x f(x)]^{\prime \prime \prime} &=x f^{\prime \prime \prime}(x)+f^{\prime \prime}(x)+2 f^{\prime \prime}(x) \\
& =x f^{\prime \prime \prime}(x)+3 f^{\prime \prime}(x),
\end{aligned}
$$
and so on. In general,we get:
$$
[x f(x)]^{(n)}=x f^{(n)}(x)+n f^{(n-1)}(x).
$$
