u ′ ⋅ v + u ⋅ v ′ or in Leibniz's notation as d d x ( u ⋅ v )d u d x ⋅ v + u ⋅ d v d x .
v d u + u d v dan ini merupakan bentuk diferensial dari kaidah darab.
− u ⋅ vu ⋅ d v + v ⋅ d u + d u ⋅ d v .
( u + d u ) ( v + d v ) − u vu d v + v d u + d u d v .
d(u\cdot v)&{}=(u+du)\cdot (v+dv)-u\cdot v\\&{}=u\cdot dv+v\cdot du+du\cdot dv.\end{aligned}}} Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that d ( u ⋅ v )v ⋅ d u + u ⋅ d v and this is indeed the differential form of the product rule.
d(uv)&{}=(u+du)(v+dv)-uv\\&{}=u\,dv+v\,du+du\,dv.\end{aligned}}} Oleh karena (du)(dv) adalah "dapat diabaikan" (i.e. paling tidak kuadratis pada du dan dv), Leibniz berkesimpula bahwa d ( u v )
If we divide through by the differential dx, we obtain d d x ( u ⋅ v )
Jika kita membaginya dengan dx, kita mendapatkan d d x ( u v )
v ⋅ d u d x + u ⋅ d v d x {dx}}(u\cdot v)=v\cdot {\frac {du}{dx}}+u\cdot {\frac {dv}{dx}}} which can also be written in Lagrange's notation as ( u ⋅ v ) ′v ⋅ u ′ + u ⋅ v ′ .
v ( d u d x ) + u ( d v d x ) {dx}}(uv)=v\left({\frac {du}{dx}}\right)+u\left({\frac {dv}{dx}}\right)} yang dapat ditulis dengan "notasi prima" sebagai ( u v )
− f ( x ) Δ x ⋅ lim Δ x → 0 g ( x + Δ x ) + lim Δ x → 0 f ( x ) ⋅ lim Δ x → 0 g ( x + Δ x ) − g ( x ) Δ xf ′ ( x )
− h ( x ) w − xlim w → x f ( w ) g ( w ) − f ( x ) g ( x ) w − x .
g ( x ) + f ( x )
g ′ ( x ) + g ( x ) f ′ ( x ) .
v u ′ + u v ′ .
′v u ′ + u v ′ .
d x∂ ( u v ) ∂ u d u d x + ∂ ( u v ) ∂ v d v d xv d u d x + u d v d x .
d xd u d x v w + u d v d x w + u v d w d x .
={\frac {\partial (uv)}{\partial u}}{\frac {du}{dx}}+{\frac {\partial (uv)}{\partial v}}{\frac {dv}{dx}}=v{\frac {du}{dx}}+u{\frac {dv}{dx}}.}
={\frac {\partial (ab)}{\partial a}}{\frac {da}{dx}}+{\frac {\partial (ab)}{\partial b}}{\frac {db}{dx}}=b{\frac {da}{dx}}+a{\frac {db}{dx}}.}
d xst ⁡ ( ( u + d u ) ( v + d v )
u d v d x + v d u d x .
Then d uu ′ d x and d vv ′ d x , so that d ( u v )( u + d u ) ( v + d v )
v d u + u d v dan ini merupakan bentuk diferensial dari kaidah darab.
′u ⋅ v ′ + v ⋅ u ′ .
′v u ′ + u v ′ .
f ′ ( x ) f ( x ) + g ′ ( x )
g ′ ( x ) + g ( x ) f ′ ( x ) .
g ( x ) {h(x)}}={\frac {f'(x)}{f(x)}}+{\frac {g'(x)}{g(x)}}} Solving for h ′ ( x ) and substituting back f ( x ) g ( x ) for h ( x ) gives: h ′ ( x )
g ( x ) , dan f and g masing-masing terdiferensialkan pada bilangan tetap x. Maka h ′ ( x )lim w → x h ( w )
h ( x ) ( f ′ ( x ) f ( x ) + g ′ ( x )
− f ( x ) w − x ) ) .
g ( x ) )f ( x ) g ( x ) ( f ′ ( x ) f ( x ) + g ′ ( x )
g ′ ( x ) + g ( x ) f ′ ( x ) .
g ( x ) )f ′ ( x ) g ( x ) + f ( x )
g ′ ( x ) + g ( x ) f ′ ( x ) .
For example, for three factors we have d ( u v w ) d xd u d x v w + u d v d x w + u v d w d x . {dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}.} For a collection of functions f 1 , … , f k ,\dots ,f_{k}} , we have d d x [ ∏ i1 k f i ( x ) ]∑ i1 k ( ( d d x f i ( x ) ) ∏ j1 , j ≠ i k f j ( x ) )
{dx}}={\frac {du}{dx}}vw+u{\frac {dv}{dx}}w+uv{\frac {dw}{dx}}.} Untuk sekumpulan fungsi f 1 , … , f k ,\dots ,f_{k}} : d d x ∏ i1 k f i ( x )( ∑ i1 k d d x f i ( x ) f i ( x ) ) ∏ i1 k f i ( x ) .
{dx}}\left[\prod _{i=1}^{k}f_{i}(x)\right]=\sum _{i=1}^{k}\left(\left({\frac {d}{dx}}f_{i}(x)\right)\prod _{j=1,j\neq i}^{k}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).}
{dx}}\prod _{i=1}^{k}f_{i}(x)=\left(\sum _{i=1}^{k}{\frac {{\frac {d}{dx}}f_{i}(x)}{f_{i}(x)}}\right)\prod _{i=1}^{k}f_{i}(x).}
d n ( u v )∑ k0 n ( n k ) ⋅ d ( n − k ) ( u ) ⋅ d ( k ) ( v ) .
∑ k0 n ( n k ) u ( n − k ) ( x ) v ( k ) ( x ) .
∑ k0 n ( n k ) ⋅ u ( n − k ) ( x ) ⋅ v ( k ) ( x ) .
∑ k0 n ( n k ) u ( n − k ) ( x ) v ( k ) ( x ) .
(x)=\sum _{k=0}^{n}{n \choose k}\cdot u^{(n-k)}(x)\cdot v^{(k)}(x).}
(x)=\sum _{k=0}^{n}{n \choose k}u^{(n-k)}(x)\;v^{(k)}(x).}
^{k}f_{i}\right)^{\!\!(n)}=\sum _{j_{1}+j_{2}+\cdots +j_{k}=n}{n \choose j_{1},j_{2},\ldots ,j_{k}}\prod _{i=1}^{k}f_{i}^{(j_{i})}.}
{dx}}\prod _{i=1}^{k}f_{i}(x)=\left(\sum _{i=1}^{k}{\frac {{\frac {d}{dx}}f_{i}(x)}{f_{i}(x)}}\right)\prod _{i=1}^{k}f_{i}(x).}
∑ S ∂ | S | u ∏ i ∈ S ∂ x i ⋅ ∂ n − | S | v ∏ i ∉ S ∂ x i \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 the index S runs through all 2n subsets of {1, ..., n}, and |S| is the cardinality of S. For example, when n3, ∂ 3 ∂ x 1 ∂ x 2 ∂ x 3 ( u v )
Untuk turunan parsial lebih tinggi: ∂ n ∂ x 1 ⋯ ∂ x n ( u v )∑ S ∂ | S | u ∏ i ∈ S ∂ x i ⋅ ∂ n − | S | v ∏ i ∉ S ∂ x i \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}}} dengan indeks S adalah deret 2n dari subhimpunan dari {1, ..., n}.
&{\partial ^{3} \over \partial x_{1}\,\partial x_{2}\,\partial x_{3}}(uv)\\[1ex]={}&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}}\\[1ex]&+{\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.\\[-3ex]&\end{aligned}}} Suppose X, Y, and Z are Banach spaces (which includes Euclidean space) and B :
&{}\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}}}Kaidah darab pada
× Y → Z is a continuous bilinear operator.
× Y → Z adalah operator bilinear kontinu.
× Y is the linear map D(x,y)B :
× Y adalah peta linear D(x,y)B: X
× Y → Z given by ( D ( x , y ) B ) ( u , v )
× Y → Z ( D ( x , y ) B ) ( u , v )B ( u , y ) + B ( x , v ) ∀ ( u , v ) ∈ X × Y .
B ( u , y ) + B ( x , v ) ∀ ( u , v ) ∈ X × Y .
× Y → Z ( D ( x , y ) B ) ( u , v )B ( u , y ) + B ( x , v ) ∀ ( u , v ) ∈ X × Y .
v ( f ) g ( p ) + f ( p ) v ( g ) .
− g ( x ) ) + g ( w ) ( f ( w )