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: $\frac{d^{n}}{d x^{n}} [f (x) g (x)] \|_{x = x_{0}} = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x_{0}) g^{(n - k)} (x_{0})$ Here, $f^{(k)}$ denotes the $k^{t h}$ derivative of $f$ (with $f^{(0)} = f, f^{(1)} = f^{'}$ , etc.), $g^{(n - k)}$ denotes the $(n - k)^{t h}$ 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: $\frac{d^{n}}{d x^{n}} [f (x) g (x)] = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x) g^{(n - k)} (x)$
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 $\frac{d^{n} (u v)}{(d x)^{n}} = \sum_{k = 0}^{n} (\binom{n}{k}) \frac{d^{k} u}{(d x)^{k}} \frac{d^{n - k} v}{(d x)^{n - k}}$

Particular cases

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