General Leibniz rule
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}} In calculus, the general Leibniz rule,<ref>Template:Cite book</ref> named after Gottfried Wilhelm Leibniz, generalizes the product rule for the derivative of the product of two functions (which is also known as "Leibniz's rule"). It states that if <math>f</math> and <math>g</math> are Template:Mvar-times differentiable functions, then the product <math>fg</math> is also Template:Mvar-times differentiable and its Template:Mvar-th derivative is given by <math display="block">(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, Template:Math, the rule gives an expression for the second derivative of a product of two functions: <math display="block">(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 f1,...,fm. <math display="block">\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 (k1,...,km) of non-negative integers with <math display="inline">\sum_{t=1}^m k_t=n,</math> and <math display="block"> {n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}</math> are the multinomial coefficients. This is akin to the multinomial formula from algebra.
Proof
The proof of the general Leibniz rule<ref name="Spivey">Template:Cite book</ref>Template:Rp 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 display="block"> (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 display="block"> (fg)^{(n)}=\sum_{k=0}^n\binom{n}{k} f^{(n-k)}g^{(k)}. </math>
Then, <math display="block">\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 Template:Nowrap and the proof is complete.
Relationship to the binomial theorem
The Leibniz rule bears a strong resemblance to the binomial theorem, and in fact the binomial theorem can be proven directly from the Leibniz rule by taking <math>f(x) = e^{ax}</math> and <math>g(x) = e^{bx},</math> which gives
- <math>(a + b)^n e^{(a+b)x} = e^{(a+b)x}\sum_{k=0}^n \binom{n}{k} a^{n-k}b^k,</math>
and then dividing both sides by <math>e^{(a+b)x}.</math><ref name="Spivey" />Template:Rp
Multivariable calculus
With the multi-index notation for partial derivatives of functions of several variables, the Leibniz rule states more generally: <math display="block">\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 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 display="block">R(x, \xi) = e^{-{\langle x, \xi \rangle}} R (e^{\langle x, \xi \rangle}).</math>
A direct computation now gives: <math display="block">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.