Product rule for higher derivatives

Statement

This states that if $f$ and $g$ are $n$ times differentiable functions at $x=x_0$, then the pointwise product $f\cdot g$ is also $n$ times differentiable at $x=x_0$, and we have:

$\frac{d^n}{d{x}^n}[f(x)g(x)]{|}_{x=x_0}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}(x_0)g^{(n-k)}(x_0)$

Here, $f^{(k)}$ denotes the $k^{th}$ derivative of $f$, $g^{(n-k)}$ denotes the $(n-k{)}^{th}$ derivative of $g$, and $(\genfrac{}{}{0ex}{}{n}{k})$ is the binomial coefficient.

If we consider this as a general expression rather than evaluating at a given point, we get:

$\frac{d^n}{d{x}^n}[f(x)g(x)]=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}(x)g^{(n-k)}(x)$

Particular cases

Value of $n$	Formula for $\frac{d^n}{d{x}^n}[f(x)g(x)]$
1	$f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$ (this is the usual product rule for differentiation).
2	$f^{″}(x)g(x)+2f^{\prime }(x)g^{\prime }(x)+f(x)g^{″}(x)$.
3	$f^{‴}(x)g(x)+3f^{″}(x)g^{\prime }(x)+3f^{\prime }(x)g^{″}(x)+g^{‴}(x)$.