Product rule for higher derivatives

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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 ${\frac {d^{n}(uv)}{(dx)^{n}}}=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {d^{n-k}u}{(dx)^{n-k}}}{\frac {d^{k}v}{(dx)^{k}}}$

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 $(f\cdot g)^{(n)}$.