Product rule for higher derivatives

This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
View other differentiation rules

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:

${\frac {d^{n}}{dx^{n}}}[f(x)g(x)]|_{x=x_{0}}=\sum _{k=0}^{n}{\binom {n}{k}}f^{(k)}(x_{0})g^{(n-k)}(x_{0})$

Here, $f^{(k)}$ denotes the $k^{th}$ derivative of $f$, $g^{(n-k)}$ denotes the $(n-k)^{th}$ derivative of $g$, and ${\binom {n}{k}}$ is the binomial coefficient. These are the same as the coefficients that appear in the expansion of $\!(A+B)^{n}$.

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

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

Particular cases

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