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

Version type	Statement
specific point, named functions	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 $f$ (with $f^{(0)}=f,f^{(1)}=f'$, etc.), $g^{(n-k)}$ 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 $\!(A+B)^{n}$.
generic point, named functions, point notation	If $f$ and $g$ are functions of one variable, the following holds wherever the right side makes sense: $\!{\frac {d^{n}}{dx^{n}}}[f(x)g(x)]=\sum _{k=0}^{n}{\binom {n}{k}}f^{(k)}(x)g^{(n-k)}(x)$
generic point, named functions, point-free notation	If $f$ and $g$ are functions of one variable, the following holds wherever the right side makes sense: $\!(f\cdot g)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}f^{(k)}g^{(n-k)}$
Pure Leibniz notation	Suppose $u$ and $v$ are both variables functionally dependent on $x$. Then ${\frac {d^{n}(uv)}{(dx)^{n}}}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {d^{k}u}{(dx)^{k}}}{\frac {d^{n-k}v}{(dx)^{n-k}}}$

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	$\!f'''(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x)+g'''(x)$.
4	$\!f^{(4)}(x)g(x)+4f'''(x)g'(x)+6f''(x)g''(x)+4f'(x)g'''(x)+f(x)g^{(4)}(x)$
5	$\!f^{(5)}(x)g(x)+5f^{(4)}(x)g'(x)+10f'''(x)g''(x)+10f''(x)g'''(x)+5f(x)g^{(4)}(x)+g^{(5)}(x)$