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In calculus , the general Leibniz rule ,[ 1] Gottfried Wilhelm Leibniz , generalizes the product rule (which is also known as "Leibniz's rule"). It states that if
f
{\displaystyle f}
g
{\displaystyle g}
n
{\displaystyle n}
differentiable functions , then the product
f
g
{\displaystyle fg}
n
{\displaystyle n}
n
{\displaystyle n}
(
f
g
)
(
n
)
(
x
)
=
∑
k
=
0
n
(
n
k
)
f
(
n
−
k
)
(
x
)
g
(
k
)
(
x
)
{\displaystyle (fg)^{(n)}(x)=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}(x)g^{(k)}(x)}
where
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
{\displaystyle {n \choose k}={n! \over k!(n-k)!}}
binomial coefficient and
f
(
0
)
(
x
)
=
f
(
x
)
{\displaystyle f^{(0)}(x)=f(x)}
This can be proved by using the product rule and mathematical induction (see proof below).
Second derivative
In case
n
=
2
{\displaystyle n=2}
(
f
g
)
″
(
x
)
=
∑
k
=
0
2
(
2
k
)
f
(
2
−
k
)
(
x
)
g
(
k
)
(
x
)
=
f
″
(
x
)
g
(
x
)
+
2
f
′
(
x
)
g
′
(
x
)
+
f
(
x
)
g
″
(
x
)
.
{\displaystyle (fg)''(x)=\sum \limits _{k=0}^{2}{{2 \choose k}f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).}
The binomial coefficients can be deduced thanks to the Pascal's triangle .
More than two factors
The formula can be generalized to the product of m differentiable functions f 1 ,...,f m
(
f
1
f
2
⋯
f
m
)
(
n
)
=
∑
k
1
+
k
2
+
⋯
+
k
m
=
n
(
n
k
1
,
k
2
,
…
,
k
m
)
∏
1
≤
t
≤
m
f
t
(
k
t
)
,
{\displaystyle \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\leq t\leq m}f_{t}^{(k_{t})}\,,}
where the sum extends over all m -tuples (k 1 ,...,k m
∑
t
=
1
m
k
t
=
n
{\displaystyle \sum _{t=1}^{m}k_{t}=n}
(
n
k
1
,
k
2
,
…
,
k
m
)
=
n
!
k
1
!
k
2
!
⋯
k
m
!
{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}}
are the multinomial coefficients . This is akin to the multinomial formula from algebra.
Proof
Proof
Show that the equality holds for any functions
f
{\displaystyle f}
g
{\displaystyle g}
n
{\displaystyle n}
Basis
At rank
m
=
1
{\displaystyle m=1}
(
f
g
)
′
(
x
)
=
f
′
(
x
)
g
(
x
)
+
f
(
x
)
g
′
(
x
)
{\displaystyle (fg)'(x)=f'(x)g(x)+f(x)g'(x)}
and
∑
k
=
0
1
(
1
k
)
f
(
1
−
k
)
(
x
)
g
(
k
)
(
x
)
=
f
′
(
x
)
g
(
x
)
+
f
(
x
)
g
′
(
x
)
{\displaystyle \sum _{k=0}^{1}{1 \choose k}f^{(1-k)}(x)g^{(k)}(x)=f'(x)g(x)+f(x)g'(x)}
Hence, the equality holds at the initial rank.
Inductive step
We assume that the equality
(
f
g
)
(
m
)
(
x
)
=
∑
k
=
0
m
(
m
k
)
f
(
m
−
k
)
(
x
)
g
(
k
)
(
x
)
{\displaystyle (fg)^{(m)}(x)=\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k)}(x)}
holds for
m
∈
N
∗
{\displaystyle m\in \mathbb {N} ^{*}}
Therefore, at rank
(
m
+
1
)
{\displaystyle (m+1)}
(
f
g
)
(
m
+
1
)
(
x
)
=
d
d
x
∑
k
=
0
m
(
m
k
)
f
(
m
−
k
)
(
x
)
g
(
k
)
(
x
)
=
∑
k
=
0
m
(
m
k
)
f
(
m
−
k
)
(
x
)
g
(
k
+
1
)
(
x
)
+
∑
k
=
0
m
(
m
k
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
=
∑
k
=
1
m
+
1
(
m
k
−
1
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
+
∑
k
=
0
m
(
m
k
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
=
(
m
m
)
f
(
x
)
g
(
m
+
1
)
(
x
)
+
∑
k
=
1
m
(
m
k
−
1
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
+
∑
k
=
1
m
(
m
k
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
+
(
m
0
)
f
(
m
+
1
)
(
x
)
g
(
x
)
=
(
m
m
)
f
(
x
)
g
(
m
+
1
)
(
x
)
+
∑
k
=
1
m
[
(
m
k
−
1
)
+
(
m
k
)
]
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
+
(
m
0
)
f
(
m
+
1
)
(
x
)
g
(
x
)
=
(
m
+
1
m
+
1
)
f
(
x
)
g
(
m
+
1
)
(
x
)
+
∑
k
=
1
m
(
m
+
1
k
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
+
(
m
+
1
0
)
f
(
m
+
1
)
(
x
)
g
(
x
)
=
∑
k
=
0
m
+
1
(
m
+
1
k
)
f
(
m
+
1
−
k
)
(
x
)
g
(
k
)
(
x
)
.
{\displaystyle {\begin{aligned}(fg)^{(m+1)}(x)&={\frac {\text{d}}{{\text{d}}x}}\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k)}(x)\\&=\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k+1)}(x)+\sum _{k=0}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)\\&=\sum _{k=1}^{m+1}{{m} \choose {k-1}}f^{(m+1-k)}(x)g^{(k)}(x)+\sum _{k=0}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)\\&={{m} \choose {m}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}{{m} \choose {k-1}}f^{(m+1-k)}(x)g^{(k)}(x)+\sum _{k=1}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)+{{m} \choose {0}}f^{(m+1)}(x)g(x)\\&={{m} \choose {m}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}\left[{{m} \choose {k-1}}+{{m} \choose {k}}\right]f^{(m+1-k)}(x)g^{(k)}(x)+{{m} \choose {0}}f^{(m+1)}(x)g(x)\\&={{m+1} \choose {m+1}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}{{m+1} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)+{{m+1} \choose {0}}f^{(m+1)}(x)g(x)\\&=\sum _{k=0}^{m+1}{{m+1} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x).\end{aligned}}}
Multivariable calculus
With the multi-index notation for partial derivatives of functions of several variables, the Leibniz rule states more generally:
∂
α
(
f
g
)
=
∑
{
β
:
β
≤
α
}
(
α
β
)
(
∂
β
f
)
(
∂
α
−
β
g
)
.
{\displaystyle \partial ^{\alpha }(fg)=\sum _{\{\beta \,:\,\beta \leq \alpha \}}{\alpha \choose \beta }(\partial ^{\beta }f)(\partial ^{\alpha -\beta }g).}
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
R
=
P
∘
Q
{\displaystyle R=P\circ Q}
R is also a differential operator, the symbol of R is given by:
R
(
x
,
ξ
)
=
e
−
⟨
x
,
ξ
⟩
R
(
e
⟨
x
,
ξ
⟩
)
.
{\displaystyle R(x,\xi )=e^{-{\langle x,\xi \rangle }}R(e^{\langle x,\xi \rangle }).}
A direct computation now gives:
R
(
x
,
ξ
)
=
∑
α
1
α
!
(
∂
∂
ξ
)
α
P
(
x
,
ξ
)
(
∂
∂
x
)
α
Q
(
x
,
ξ
)
.
{\displaystyle 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 ).}
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
References
![{\displaystyle {\begin{aligned}(fg)^{(m+1)}(x)&={\frac {\text{d}}{{\text{d}}x}}\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k)}(x)\\&=\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k+1)}(x)+\sum _{k=0}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)\\&=\sum _{k=1}^{m+1}{{m} \choose {k-1}}f^{(m+1-k)}(x)g^{(k)}(x)+\sum _{k=0}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)\\&={{m} \choose {m}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}{{m} \choose {k-1}}f^{(m+1-k)}(x)g^{(k)}(x)+\sum _{k=1}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)+{{m} \choose {0}}f^{(m+1)}(x)g(x)\\&={{m} \choose {m}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}\left[{{m} \choose {k-1}}+{{m} \choose {k}}\right]f^{(m+1-k)}(x)g^{(k)}(x)+{{m} \choose {0}}f^{(m+1)}(x)g(x)\\&={{m+1} \choose {m+1}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}{{m+1} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)+{{m+1} \choose {0}}f^{(m+1)}(x)g(x)\\&=\sum _{k=0}^{m+1}{{m+1} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eaf0c781666a44f905642afb7965a056697c979e)
