Generalization of the product rule in calculus
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 -times differentiable functions , then the product
f
g
{\displaystyle fg}
n -times differentiable and its n -th derivative is given by
(
f
g
)
(
n
)
=
∑
k
=
0
n
(
n
k
)
f
(
n
−
k
)
g
(
k
)
,
{\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{(k)},}
where
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
{\displaystyle {n \choose k}={n! \over k!(n-k)!}}
is the
binomial coefficient and
f
(
j
)
{\displaystyle f^{(j)}}
denotes the
j th derivative of
f (and in particular
f
(
0
)
=
f
{\displaystyle f^{(0)}=f}
).
The rule can be proven by using the product rule and mathematical induction .
Second derivative If, for example, 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}{{\binom {2}{k}}f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).}
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 ) of non-negative integers with
∑
t
=
1
m
k
t
=
n
,
{\textstyle \sum _{t=1}^{m}k_{t}=n,}
and
(
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 formula
from algebra.
Proof The proof of the general Leibniz rule proceeds by induction. Let
f
{\displaystyle f}
g
{\displaystyle g}
n
{\displaystyle n}
n
=
1
{\displaystyle n=1}
(
f
g
)
′
=
f
′
g
+
f
g
′
,
{\displaystyle (fg)'=f'g+fg',}
which is the usual
product rule and is known to be true. Next, assume that the statement holds for a fixed
n
≥
1
,
{\displaystyle n\geq 1,}
that is, that
(
f
g
)
(
n
)
=
∑
k
=
0
n
(
n
k
)
f
(
n
−
k
)
g
(
k
)
.
{\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k)}.}
Then,
(
f
g
)
(
n
+
1
)
=
[
∑
k
=
0
n
(
n
k
)
f
(
n
−
k
)
g
(
k
)
]
′
=
∑
k
=
0
n
(
n
k
)
f
(
n
+
1
−
k
)
g
(
k
)
+
∑
k
=
0
n
(
n
k
)
f
(
n
−
k
)
g
(
k
+
1
)
=
∑
k
=
0
n
(
n
k
)
f
(
n
+
1
−
k
)
g
(
k
)
+
∑
k
=
1
n
+
1
(
n
k
−
1
)
f
(
n
+
1
−
k
)
g
(
k
)
=
(
n
0
)
f
(
n
+
1
)
g
(
0
)
+
∑
k
=
1
n
(
n
k
)
f
(
n
+
1
−
k
)
g
(
k
)
+
∑
k
=
1
n
(
n
k
−
1
)
f
(
n
+
1
−
k
)
g
(
k
)
+
(
n
n
)
f
(
0
)
g
(
n
+
1
)
=
(
n
+
1
0
)
f
(
n
+
1
)
g
(
0
)
+
(
∑
k
=
1
n
[
(
n
k
−
1
)
+
(
n
k
)
]
f
(
n
+
1
−
k
)
g
(
k
)
)
+
(
n
+
1
n
+
1
)
f
(
0
)
g
(
n
+
1
)
=
(
n
+
1
0
)
f
(
n
+
1
)
g
(
0
)
+
∑
k
=
1
n
(
n
+
1
k
)
f
(
n
+
1
−
k
)
g
(
k
)
+
(
n
+
1
n
+
1
)
f
(
0
)
g
(
n
+
1
)
=
∑
k
=
0
n
+
1
(
n
+
1
k
)
f
(
n
+
1
−
k
)
g
(
k
)
.
{\displaystyle {\begin{aligned}(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{aligned}}}
And so the statement holds for
n
+
1
{\displaystyle n+1}
and the proof is complete.
Multivariable calculus With the
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.}
Since
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)^{(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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/195166910864530fea5b647df661b47df305b4b3)
