Generalizations Product rule

1 generalizations

1.1 product of more 2 factors
1.2 higher derivatives
1.3 higher partial derivatives
1.4 banach space
1.5 derivations in abstract algebra
1.6 vector functions
1.7 scalar fields

generalizations

a product of more 2 factors

the product rule can generalized products of more 2 factors. example, 3 factors have

d
(
u
v
w
)


d
x



=



d
u


d
x



v
w
+
u



d
v


d
x



w
+
u
v



d
w


d
x





{\displaystyle {\frac {d(uvw)}{dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}}

.

for collection of functions




f

1


,
…
,

f

k




{\displaystyle f_{1},\dots ,f_{k}}

, have

d

d
x




[

∏

i
=
1


k



f

i


(
x
)
]

=

∑

i
=
1


k



(


d

d
x




f

i


(
x
)

∏

j
≠
i



f

j


(
x
)
)

=

(

∏

i
=
1


k



f

i


(
x
)
)


(

∑

i
=
1


k






f

i

′

(
x
)



f

i


(
x
)



)

.


{\displaystyle {\frac {d}{dx}}\left[\prod _{i=1}^{k}f_{i}(x)\right]=\sum _{i=1}^{k}\left({\frac {d}{dx}}f_{i}(x)\prod _{j\neq i}f_{j}(x)\right)=\left(\prod _{i=1}^{k}f_{i}(x)\right)\left(\sum _{i=1}^{k}{\frac {f _{i}(x)}{f_{i}(x)}}\right).}



higher derivatives

it can generalized general leibniz rule nth derivative of product of 2 factors, symbolically expanding according binomial theorem:

d

n


(
u
v
)
=

∑

k
=
0


n





(


n
k


)



⋅

d

(
n
−
k
)


(
u
)
⋅

d

(
k
)


(
v
)
.


{\displaystyle d^{n}(uv)=\sum _{k=0}^{n}{n \choose k}\cdot d^{(n-k)}(u)\cdot d^{(k)}(v).}

applied @ specific point x, above formula gives:

(
u
v

)

(
n
)


(
x
)
=

∑

k
=
0


n





(


n
k


)



⋅

u

(
n
−
k
)


(
x
)
⋅

v

(
k
)


(
x
)
.


{\displaystyle (uv)^{(n)}(x)=\sum _{k=0}^{n}{n \choose k}\cdot u^{(n-k)}(x)\cdot v^{(k)}(x).}

furthermore, nth derivative of arbitrary number of factors:

(

∏

i
=
1


k



f

i


)


(
n
)


=

∑


j

1


+

j

2


+
.
.
.
+

j

k


=
n





(


n


j

1


,

j

2


,
.
.
.
,

j

k





)




∏

i
=
1


k



f

i


(

j

i


)


.


{\displaystyle \left(\prod _{i=1}^{k}f_{i}\right)^{(n)}=\sum _{j_{1}+j_{2}+...+j_{k}=n}{n \choose j_{1},j_{2},...,j_{k}}\prod _{i=1}^{k}f_{i}^{(j_{i})}.}



higher partial derivatives

for partial derivatives, have

∂

n



∂

x

1



⋯

∂

x

n





(
u
v
)
=

∑

s






∂


|

s

|



u



∏

i
∈
s


∂

x

i





⋅




∂

n
−

|

s

|



v



∏

i
∉
s


∂

x

i







{\displaystyle {\partial ^{n} \over \partial x_{1}\,\cdots \,\partial x_{n}}(uv)=\sum _{s}{\partial ^{|s|}u \over \prod _{i\in s}\partial x_{i}}\cdot {\partial ^{n-|s|}v \over \prod _{i\not \in s}\partial x_{i}}}

where index s runs through whole list of 2 subsets of {1, ..., n}. example, when n = 3, then

∂

3



∂

x

1



∂

x

2



∂

x

3





(
u
v
)












=
u
⋅




∂

3


v


∂

x

1



∂

x

2



∂

x

3





+



∂
u


∂

x

1





⋅




∂

2


v


∂

x

2



∂

x

3





+



∂
u


∂

x

2





⋅




∂

2


v


∂

x

1



∂

x

3





+



∂
u


∂

x

3





⋅




∂

2


v


∂

x

1



∂

x

2


















+




∂

2


u


∂

x

1



∂

x

2





⋅



∂
v


∂

x

3





+




∂

2


u


∂

x

1



∂

x

3





⋅



∂
v


∂

x

2





+




∂

2


u


∂

x

2



∂

x

3





⋅



∂
v


∂

x

1





+




∂

3


u


∂

x

1



∂

x

2



∂

x

3





⋅
v
.






{\displaystyle {\begin{aligned}&{}\quad {\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}(uv)\\\\&{}=u\cdot {\partial ^{3}v \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{1}}\cdot {\partial ^{2}v \over \partial x_{2}\,\partial x_{3}}+{\partial u \over \partial x_{2}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{3}}+{\partial u \over \partial x_{3}}\cdot {\partial ^{2}v \over \partial x_{1}\,\partial x_{2}}\\\\&{}\qquad +{\partial ^{2}u \over \partial x_{1}\,\partial x_{2}}\cdot {\partial v \over \partial x_{3}}+{\partial ^{2}u \over \partial x_{1}\,\partial x_{3}}\cdot {\partial v \over \partial x_{2}}+{\partial ^{2}u \over \partial x_{2}\,\partial x_{3}}\cdot {\partial v \over \partial x_{1}}+{\partial ^{3}u \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}\cdot v.\end{aligned}}}



banach space

suppose x, y, , z banach spaces (which includes euclidean space) , b : x × y → z continuous bilinear operator. b differentiable, , derivative @ point (x,y) in x × y linear map d(x,y)b : x × y → z given by

(

d


(
x
,
y
)




b
)

(
u
,
v
)

=
b

(
u
,
y
)

+
b

(
x
,
v
)


∀
(
u
,
v
)
∈
x
×
y
.


{\displaystyle (d_{\left(x,y\right)}\,b)\left(u,v\right)=b\left(u,y\right)+b\left(x,v\right)\qquad \forall (u,v)\in x\times y.}



derivations in abstract algebra

in abstract algebra, product rule used define called derivation, not vice versa.

vector functions

the product rule extends scalar multiplication, dot products, , cross products of vector functions.

for scalar multiplication:



(
f
⋅


g



)
′

=
f


′

⋅


g


+
f
⋅


g




′



{\displaystyle (f\cdot {\mathbf {g}}) =f\; \cdot {\mathbf {g}}+f\cdot {\mathbf {g}}\; }

for dot products:



(


f


⋅


g



)
′

=


f




′

⋅


g


+


f


⋅


g




′



{\displaystyle ({\mathbf {f}}\cdot {\mathbf {g}}) ={\mathbf {f}}\; \cdot {\mathbf {g}}+{\mathbf {f}}\cdot {\mathbf {g}}\; }

for cross products:



(


f


×


g



)
′

=


f




′

×


g


+


f


×


g




′



{\displaystyle ({\mathbf {f}}\times {\mathbf {g}}) ={\mathbf {f}}\; \times {\mathbf {g}}+{\mathbf {f}}\times {\mathbf {g}}\; }

note: cross products not commutative, i.e.



(
f
×
g

)
′

≠

f
′

×
g
+

g
′

×
f


{\displaystyle (f\times g) \neq f \times g+g \times f}

, instead products anticommutative, can written



(
f
×
g

)
′

=

f
′

×
g
−

g
′

×
f


{\displaystyle (f\times g) =f \times g-g \times f}

scalar fields

for scalar fields concept of gradient analog of derivative:

∇
(
f
⋅
g
)
=
∇
f
⋅
g
+
f
⋅
∇
g


{\displaystyle \nabla (f\cdot g)=\nabla f\cdot g+f\cdot \nabla g}