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wikipedia>Blue2718: /* Proof */ Fixed a typo in the proof. The terms k = 0 and k = n+1 should have had zeroth order derivatives for g and f respectively.

2023-07-13T15:41:25Z

Proof: Fixed a typo in the proof. The terms k = 0 and k = n+1 should have had zeroth order derivatives for g and f respectively.

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{{Short description|Generalization of the product rule in calculus}}
{{other uses|Leibniz's rule (disambiguation)}}
{{Calculus}}
In [[calculus]], the '''general Leibniz rule''',<ref>{{cite book |last=Olver |first=Peter J. |year=2000 |title=Applications of Lie Groups to Differential Equations |publisher=Springer |pages=318–319 |isbn=9780387950006 |url=https://books.google.com/books?id=sI2bAxgLMXYC&pg=PA318 }}</ref> named after [[Gottfried Wilhelm Leibniz]], generalizes the [[product rule]] (which is also known as "Leibniz's rule"). It states that if <math>f</math> and <math>g</math> are <math>n</math>-times [[differentiable function]]s, then the product <math>fg</math> is also <math>n</math>-times differentiable and its <math>n</math>th derivative is given by

:<math>(fg)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(n-k)} g^{(k)},</math>

where <math>{n \choose k}={n!\over k! (n-k)!}</math> is the [[binomial coefficient]] and <math>f^{(j)}</math> denotes the ''j''th derivative of ''f'' (and in particular <math>f^{(0)}= f</math>).

The rule can be proven by using the product rule and [[mathematical induction]].

==Second derivative==
If, for example, {{math|1=''n'' = 2}}, the rule gives an expression for the second derivative of a product of two functions:
:<math>(fg)''(x)=\sum\limits_{k=0}^{2}{\binom{2}{k} f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).</math>

==More than two factors==
The formula can be generalized to the product of ''m'' differentiable functions ''f''1,...,''f''''m''.
:<math>\left(f_1 f_2 \cdots f_m\right)^{(n)}=\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
\prod_{1\le t\le m}f_{t}^{(k_{t})}\,,</math>
where the sum extends over all ''m''-tuples (''k''1,...,''k''''m'') of non-negative integers with <math>\sum_{t=1}^m k_t=n,</math> and
:<math> {n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}</math>
are the [[multinomial coefficient]]s. This is akin to the [[multinomial formula]] from algebra.

==Proof==

The proof of the general Leibniz rule proceeds by induction. Let <math>f</math> and <math>g</math> be <math>n</math>-times differentiable functions. The base case when <math>n=1</math> claims that:

:<math> (fg)'=f'g+fg',</math>

which is the usual product rule and is known to be true. Next, assume that the statement holds for a fixed <math>n \geq 1,</math> that is, that

:<math> (fg)^{(n)}=\sum_{k=0}^n\binom{n}{k} f^{(n-k)}g^{(k)}. </math>

Then,

:<math>\begin{align}
(fg)^{(n+1)} &= \left[ \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)} \right]' \\
&= \sum_{k=0}^n \binom{n}{k} f^{(n+1-k)} g^{(k)} + \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k+1)} \\
&= \sum_{k=0}^n \binom{n}{k} f^{(n+1-k)} g^{(k)} + \sum_{k=1}^{n+1} \binom{n}{k-1} f^{(n+1-k)} g^{(k)} \\
&= \binom{n}{0} f^{(n+1)} g^{(0)} + \sum_{k=1}^{n} \binom{n}{k} f^{(n+1-k)} g^{(k)} + \sum_{k=1}^n \binom{n}{k-1} f^{(n+1-k)} g^{(k)} + \binom{n}{n} f^{(0)} g^{(n+1)} \\
&= \binom{n+1}{0} f^{(n+1)} g^{(0)} + \left( \sum_{k=1}^n \left[\binom{n}{k-1} + \binom{n}{k} \right]f^{(n+1-k)} g^{(k)} \right) + \binom{n+1}{n+1} f^{(0)} g^{(n+1)} \\
&= \binom{n+1}{0} f^{(n+1)} g^{(0)} + \sum_{k=1}^n \binom{n+1}{k} f^{(n+1-k)} g^{(k)} + \binom{n+1}{n+1}f^{(0)} g^{(n+1)} \\
&= \sum_{k=0}^{n+1} \binom{n+1}{k} f^{(n+1-k)} g^{(k)} .
\end{align}</math>

And so the statement holds for <math>n+1,</math> and the proof is complete.

==Multivariable calculus==
With the [[multi-index]] notation for [[partial derivative]]s of functions of several variables, the Leibniz rule states more generally:

:<math>\partial^\alpha (fg) = \sum_{ \beta\,:\,\beta \le \alpha } {\alpha \choose \beta} (\partial^{\beta} f) (\partial^{\alpha - \beta} g).</math>

This formula can be used to derive a formula that computes the [[symbol of a differential operator|symbol]] of the composition of differential operators. In fact, let ''P'' and ''Q'' be differential operators (with coefficients that are differentiable sufficiently many times) and <math>R = P \circ Q.</math> Since ''R'' is also a differential operator, the symbol of ''R'' is given by:

:<math>R(x, \xi) = e^{-{\langle x, \xi \rangle}} R (e^{\langle x, \xi \rangle}).</math>

A direct computation now gives:

:<math>R(x, \xi) = \sum_\alpha {1 \over \alpha!} \left({\partial \over \partial \xi}\right)^\alpha P(x, \xi) \left({\partial \over \partial x}\right)^\alpha Q(x, \xi).</math>

This formula is usually known as the Leibniz formula. It is used to define the composition in the space of symbols, thereby inducing the ring structure.

==See also==

* {{annotated link|Binomial theorem}}
* {{annotated link|Derivation (differential algebra)}}
* {{annotated link|Derivative}}
* {{annotated link|Differential algebra}}
* {{annotated link|Pascal's triangle}}
* {{annotated link|Product rule}}
* {{annotated link|Quotient rule}}

==References==
{{reflist}}

{{Calculus topics}}

[[Category:Articles containing proofs]]
[[Category:Differentiation rules]]
[[Category:Gottfried Wilhelm Leibniz]]
[[Category:Mathematical identities]]
[[Category:Theorems in analysis]]
[[Category:Theorems in calculus]]

General Leibniz rule - Revision history

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wikipedia>Blue2718: /* Proof */ Fixed a typo in the proof. The terms k = 0 and k = n+1 should have had zeroth order derivatives for g and f respectively.