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: $\frac{d^n}{dx^n}[f(x)g(x)]|_{x = x_0} = \sum_{k=0}^n \binom{n}{k} f^{(n - k)}(x_0)g^{(k)}(x_0)$ Here, $f^{(n - k)}$ denotes the $(n-k)^{th}$ derivative of $f$ (with $f^{(0)} = f, f^{(1)} = f'$, etc.), $g^{(k)}$ denotes the $k^{th}$ 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: $\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)$
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: $\! (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}$
Pure Leibniz notation	Suppose $u$ and $v$ are both variables functionally dependent on $x$. Then $\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}$

One-sided version

There are analogues of each of the statements with one-sided derivatives. Fill this in later

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

Related rules

Similar rules in single variable calculus

• Product rule for differentiation
• Chain rule for higher derivatives
• Chain rule for differentiation
• Repeated differentiation is linear

Similar rules in multivariable calculus

• Product rule for higher partial derivatives

Similar rules in advanced mathematics

• Binomial formula for powers of a derivation

Significance

Qualitative and existential significance

Each of the versions has its own qualitative significance:

Version type	Significance
specific point, named functions	This tells us that if $f$ and $g$ are both $n$ times differentiable at a point $x_0$, so is $f \cdot g$.
generic point, named functions, point notation	This tells us that if $f$ and $g$ are both $n$ times differentiable on an open interval, so is $f \cdot g$.
generic point, named functions, point-free notation	This shows that the way that $(f \cdot g)^{(n)}$ behaves is governed by the nature of the derivatives (up to the $n^{th}$) of $f$ and $g$. In particular, if $f^{(n)}$ and $g^{(n)}$ are both continuous functions on an interval, so is $(f \cdot g)^{(n)}$.

Computational feasibility significance

Each of the versions has its own computational feasibility significance:

Version type	Significance
specific point, named functions	This tells us that knowing the values (in the sense of numerical values) of $f,f',f'',\dots, f^{(n)}$ and $g,g',g'',\dots,g^{(n)}$ at a point $x_0$ allows us to compute the value $(f \cdot g)^{(n)}(x_0)$ by plugging into the formula and doing a bunch of multiplications and additions.
generic point, named functions	This tells us that knowledge of the generic expressions for $f,f',f'',\dots, f^{(n)}$ and $g,g',g'',\dots,g^{(n)}$ allows us to compute the generic expression for $(f \cdot g)^{(n)}$.