The chain rule may also be expressed in Leibniz's notation.
Notasi umum aturan rantai adalah karena Leibniz.
If a variable z depends on the variable y, which itself depends on the variable x (that is, y and z are dependent variables), then z depends on x as well, via the intermediate variable y.
Bila variabel z tergantung pada variabel y, yang bergantung pada variabelnya x (yaitu y dan z adalah variabel dependen), lalu z, melalui variabel perantara y, tergantung pada x demikian juga.
In integration, the counterpart to the chain rule is the substitution rule.
" Dalam integrasi, pasangan dari aturan rantai adalah aturan substitusi.
The chain rule seems to have first been used by Gottfried Wilhelm Leibniz.
Aturan rantai tampaknya pertama kali digunakan oleh Gottfried Wilhelm Leibniz.
He used it to calculate the derivative of a + b z + c z 2 }}} as the composite of the square root function and the function a + b z + c z 2 \!} .
Dia menggunakannya untuk menghitung turunan dari a + b z + c z 2 }}} sebagai gabungan dari fungsi akar kuadrat dan fungsi a + b z + c z 2 \!} .
He first mentioned it in a 1676 memoir (with a sign error in the calculation).
Dia pertama kali menyebutkannya dalam memoar 1676 (dengan kesalahan tanda dalam perhitungan).
The common notation of the chain rule is due to Leibniz.
Notasi umum aturan rantai adalah karena Leibniz.
Guillaume de l'Hôpital used the chain rule implicitly in his Analyse des infiniment petits.
Guillaume de l'Hôpital menggunakan aturan rantai secara implisit dalam miliknya Analyse des infiniment petits.
The chain rule does not appear in any of Leonhard Euler's analysis books, even though they were written over a hundred years after Leibniz's discovery.
Aturan rantai tidak muncul di buku analisis Leonhard Euler, meskipun mereka ditulis lebih dari seratus tahun setelah penemuan Leibniz.
If yf(u) and ug(x), then this abbreviated form is written in Leibniz notation as: d y d xd y d u ⋅ d u d x . {dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}.} The points where the derivatives are evaluated may also be stated explicitly:
Dalam notasi Leibniz, adalah: d y d xd y d u | ug ( h ( a ) ) ⋅ d u d v | vh ( a ) ⋅ d v d x | xa , {dx}}=\left.{\frac {dy}{du}}\right|_{u=g(h(a))}\cdot \left.{\frac {du}{dv}}\right|_{v=h(a)}\cdot \left.{\frac {dv}{dx}}\right|_{x=a},} atau singkatnya, d y d
d y d x | xcd y d u | ug ( c ) ⋅ d u d x | xc .
d y d xd y d u ⋅ d u d x .
{dx}}\right|_{x=c}=\left.{\frac {dy}{du}}\right|_{u=g(c)}\cdot \left.{\frac {du}{dx}}\right|_{x=c}.} Carrying the same reasoning further, given n functions f 1 , … , f n ,\ldots ,f_{n}\!} with the composite function f 1 ∘ ( f 2 ∘ ⋯ ( f n − 1 ∘ f n ) ) \circ (f_{2}\circ \cdots (f_{n-1}\circ f_{n}))\!} , if each function f i \!} is differentiable at its immediate input, then the composite function is also differentiable by the repeated application of Chain Rule, where the derivative is (in Leibniz's notation): d f 1 d xd f 1 d f 2 d f 2 d f 3 ⋯ d f n d x .
{dx}}\right|_{x=c}=\left.{\frac {dy}{du}}\right|_{u=g(c)}\cdot \left.{\frac {du}{dx}}\right|_{x=c}.} Membawa alasan yang sama lebih jauh, diberikan n fungsi f 1 , … , f n ,\ldots ,f_{n}\!} dengan fungsi komposit f 1 ∘ ( f 2 ∘ ⋯ ( f n − 1 ∘ f n ) ) \circ (f_{2}\circ \cdots (f_{n-1}\circ f_{n}))\!} , bila masing-masing fungsi f i \!} dapat dibedakan pada masukan langsungnya, maka fungsi komposit juga dapat dibedakan dengan penerapan Aturan Rantai yang berulang, di mana turunannya adalah (dalam notasi Leibniz): d f 1 d xd f 1 d f 2 d f 2 d f 3 ⋯ d f n d x .
For concreteness, consider the function ye sin ⁡ ( x 2 ) .
Untuk konkretnya, pertimbangkan fungsinya ye sin ⁡ ( x 2 ) .
e u , ug ( v )sin ⁡ v , vh ( x )x 2 .
e ue sin ⁡ ( x 2 ) , d u d vg ′ ( v )cos ⁡ vcos ⁡ ( x 2 ) , d v d
y&=f(u)=e^{u},\\u&=g(v)=\sin v,\\v&=h(x)=x^{2}.\end{aligned}}} So that yf ( g ( h ( x ) ) ) .
y&=f(u)=e^{u},\\[6pt]u&=g(v)=\sin v=\sin(x^{2}),\\[6pt]v&=h(x)=x^{2}.\end{aligned}}} Turunannya adalah:
e u , d u d vg ′ ( v )cos ⁡ v , d v d
e ue sin ⁡ ( x 2 ) , d u d vg ′ ( v )cos ⁡ vcos ⁡ ( x 2 ) , d v d
{\frac {dy}{du}}&=f'(u)=e^{u},\\{\frac {du}{dv}}&=g'(v)=\cos v,\\{\frac {dv}{dx}}&=h'(x)=2x.\end{aligned}}} The chain rule states that the derivative of their composite at the point xa is: ( f ∘ g ∘ h ) ′ ( a )f ′ ( ( g ∘ h ) ( a ) ) ⋅ ( g ∘ h )
{\frac {dy}{du}}&=f'(u)=e^{u}=e^{\sin(x^{2})},\\[6pt]{\frac {du}{dv}}&=g'(v)=\cos v=\cos(x^{2}),\\[6pt]{\frac {dv}{dx}}&=h'(x)=2x.\end{aligned}}} Kaidah rantai menyatakan bahwa turunan komposit mereka pada titik xa adalah: ( f ∘ g ∘ h ) ′ ( a )
( f ′ ∘ g ∘ h ) ( a ) ⋅ ( g ′ ∘ h ) ( a ) ⋅ h ′ ( a ) .
( f ∘ g ) ′ ( h ( a ) ) ⋅ h ′ ( a )f ′ ( g ( h ( a ) ) ) ⋅ g ′ ( h ( a ) ) ⋅ h ′ ( a ) .
{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dx}}.} The derivative function is therefore: d y d xe sin ⁡ ( x 2 ) ⋅ cos ⁡ ( x 2 ) ⋅ 2 x .
{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dx}}.} Oleh karena itu, fungsi turunannya adalah: d y d xe sin ⁡ ( x 2 ) ⋅ cos ⁡ ( x 2 ) ⋅ 2 x .
{dx}}=e^{\sin(x^{2})}\cdot \cos(x^{2})\cdot 2x.} Another way of computing this derivative is to view the composite function f ∘ g ∘ h as the composite of f ∘ g and h. Applying the chain rule in this manner would yield: ( f ∘ g ∘ h ) ′ ( a )
{dx}}=e^{\sin(x^{2})}\cdot \cos(x^{2})\cdot 2x.} Cara lain untuk menghitung turunan ini adalah dengan melihat fungsi komposit f ∘ g ∘ h sebagai gabungan dari f ∘
(f\circ g\circ h)'(a)&=(f\circ g)'(h(a))\cdot h'(a)\\&=f'(g(h(a)))\cdot g'(h(a))\cdot h'(a).\end{aligned}}} This is the same as what was computed above.
(f\circ g\circ h)'(a)&=f'((g\circ h)(a))\cdot (g\circ h)'(a)\\[10pt]&=f'((g\circ h)(a))\cdot g'(h(a))\cdot h'(a)=(f'\circ g\circ h)(a)\cdot (g'\circ h)(a)\cdot h'(a).\end{aligned}}}
This should be expected because (f ∘ g) ∘ hf ∘ (g ∘ h).
hal ini harus diharapkan karena (f ∘ g) ∘ hf ∘ (g ∘ h).
Sometimes, it is necessary to differentiate an arbitrarily long composition of the form f 1 ∘ f 2 ∘ ⋯ ∘ f n − 1 ∘ f n \circ f_{2}\circ \cdots \circ f_{n-1}\circ f_{n}\!} .
Terkadang, perlu untuk membedakan komposisi bentuk yang sangat panjang f 1 ∘ f 2 ∘ ⋯ ∘ f n − 1 ∘ f n \circ f_{2}\circ \cdots \circ f_{n-1}\circ f_{n}\!} .
In this case, define f a . .
Dalam kasus ini, definisikan f a . .
bf a ∘ f a + 1 ∘ ⋯ ∘ f b − 1 ∘ f bf_{a}\circ f_{a+1}\circ \cdots \circ f_{b-1}\circ f_{b}} where f a . .
bf a ∘ f a + 1 ∘ ⋯ ∘ f b − 1 ∘ f bf_{a}\circ f_{a+1}\circ \cdots \circ f_{b-1}\circ f_{b}} di mana f a . .
af af_{a}} and f a . .
af af_{a}} dan f a . .
b ( x )x (x)=x} when b < a .
b ( x )x (x)=x} ketika b < a .
Then the chain rule takes the form D f 1 . .
Kemudian aturan rantai terbentuk D f 1 . .
D f n∏ k1 n [ D f k ∘ f ( k + 1 ) .
D f n∏ k1 n [ D f k ∘ f ( k + 1 ) . .
. n ] Df_{1\,.\,.\,n}&=(Df_{1}\circ f_{2\,.\,.\,n})(Df_{2}\circ f_{3\,.\,.\,n})\cdots (Df_{n-1}\circ f_{n\,.\,.\,n})Df_{n}\\&=\prod _{k=1}^{n}\left[Df_{k}\circ f_{(k+1)\,.\,.\,n}\right]\end{aligned}}} or, in the Lagrange notation, f 1 . . n ′ ( x )f 1 ′ ( f 2 . .
=(Df_{1}\circ f_{2\,.\,.\,n})(Df_{2}\circ f_{3\,.\,.\,n})\cdots (Df_{n-1}\circ f_{n\,.\,.\,n})Df_{n}=\prod _{k=1}^{n}\left[Df_{k}\circ f_{(k+1)\,.\,.\,n}\right]} atau, dalam notasi Lagrange, f 1 . . n ′ ( x )f 1 ′ ( f 2 . .
f 2 ′ ( f 3 . .
n ( x ) ) f 2 ′ ( f 3 .
∏ k1 n f k ′ ( f ( k + 1 . .
f n ′ ( x )∏ k1 n f k ′ ( f ( k + 1 .
( x ) ) f_{1\,.\,.\,n}'(x)&=f_{1}'\left(f_{2\,.\,.\,n}(x)\right)\;f_{2}'\left(f_{3\,.\,.\,n}(x)\right)\cdots f_{n-1}'\left(f_{n\,.\,.\,n}(x)\right)\;f_{n}'(x)\\[1ex]&=\prod _{k=1}^{n}f_{k}'\left(f_{(k+1\,.\,.\,n)}(x)\right)\end{aligned}}} The chain rule can be used to derive some well-known differentiation rules.
. n ) ( x ) ) '(x)=f_{1}'\left(f_{2\,.\,.\,n}(x)\right)\;f_{2}'\left(f_{3\,.\,.\,n}(x)\right)\cdots f_{n-1}'\left(f_{n\,.\,.\,n}(x)\right)\;f_{n}'(x)=\prod _{k=1}^{n}f_{k}'\left(f_{(k+1\,.\,.\,n)}(x)\right)}
To see this, write the function f(x)/g(x) as the product f(x) · 1/g(x).
Untuk melihat ini, tulis fungsinya f(x)/g(x) sebagai produk f(x) · 1/g(x).
{\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)&={\frac {d}{dx}}\left(f(x)\cdot {\frac {1}{g(x)}}\right)\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left({\frac {1}{g(x)}}\right).\end{aligned}}} To compute the derivative of 1/g(x), notice that it is the composite of g with the reciprocal function, that is, the function that sends x to 1/x.
{\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)&={\frac {d}{dx}}\left(f(x)\cdot {\frac {1}{g(x)}}\right)\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left({\frac {1}{g(x)}}\right).\end{aligned}}} Untuk menghitung turunan 1/g(x), perhatikan bahwa itu adalah gabungan dari g dengan fungsi timbal balik, yaitu fungsi yang mengirim x to 1/x. Turunan dari fungsi timbal balik adalah − 1 / x 2 \!} .
By applying the chain rule, the last expression becomes: f ′ ( x ) ⋅ 1 g ( x ) + f ( x ) ⋅ ( − 1 g ( x ) 2 ⋅ g ′ ( x ) )
Dengan menerapkan aturan rantai, ekspresi terakhir menjadi: f ′ ( x ) ⋅ 1 g ( x ) + f ( x ) ⋅ ( − 1 g ( x ) 2 ⋅ g ′ ( x ) )f ′ ( x )
g ( x ) 2 , {g(x)}}+f(x)\cdot \left(-{\frac {1}{g(x)^{2}}}\cdot g'(x)\right)={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}},} which is the usual formula for the quotient rule.
g ( x ) 2 , {g(x)}}+f(x)\cdot \left(-{\frac {1}{g(x)^{2}}}\cdot g'(x)\right)={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}},} yang merupakan rumus umum untuk aturan hasil bagi.
f ′ ( g ( f ( y ) ) )
f ′ ( g ( x ) )
g ′ ( f ( y ) )1 f ′ ( y )
f ′ ( g ( x ) )
1 g ′ ( f ( y ) ) .
f ′ ( g ( x ) )
{\frac {dy}{dx}}&={\frac {dy}{du}}{\frac {du}{dx}}\\{\frac {d^{2}y}{dx^{2}}}&={\frac {d^{2}y}{du^{2}}}\left({\frac {du}{dx}}\right)^{2}+{\frac {dy}{du}}{\frac {d^{2}u}{dx^{2}}}\\{\frac {d^{3}y}{dx^{3}}}&={\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\right)^{3}+3\,{\frac {d^{2}y}{du^{2}}}{\frac {du}{dx}}{\frac {d^{2}u}{dx^{2}}}+{\frac {dy}{du}}{\frac {d^{3}u}{dx^{3}}}\\{\frac {d^{4}y}{dx^{4}}}&={\frac {d^{4}y}{du^{4}}}\left({\frac {du}{dx}}\right)^{4}+6\,{\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\right)^{2}{\frac {d^{2}u}{dx^{2}}}+{\frac {d^{2}y}{du^{2}}}\left(4\,{\frac {du}{dx}}{\frac {d^{3}u}{dx^{3}}}+3\,\left({\frac {d^{2}u}{dx^{2}}}\right)^{2}\right)+{\frac {dy}{du}}{\frac {d^{4}u}{dx^{4}}}.\end{aligned}}} One proof of the chain rule begins by defining the derivative of the composite function f ∘ g, where we take the limit of the difference quotient for f ∘ g as x approaches a: ( f ∘ g ) ′ ( a )lim x → a f ( g ( x ) ) − f ( g ( a ) )
{\frac {dy}{dx}}&={\frac {dy}{du}}{\frac {du}{dx}}\\[4pt]{\frac {d^{2}y}{dx^{2}}}&={\frac {d^{2}y}{du^{2}}}\left({\frac {du}{dx}}\right)^{2}+{\frac {dy}{du}}{\frac {d^{2}u}{dx^{2}}}\\[4pt]{\frac {d^{3}y}{dx^{3}}}&={\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\right)^{3}+3\,{\frac {d^{2}y}{du^{2}}}{\frac {du}{dx}}{\frac {d^{2}u}{dx^{2}}}+{\frac {dy}{du}}{\frac {d^{3}u}{dx^{3}}}\\[4pt]{\frac {d^{4}y}{dx^{4}}}&={\frac {d^{4}y}{du^{4}}}\left({\frac {du}{dx}}\right)^{4}+6\,{\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\right)^{2}{\frac {d^{2}u}{dx^{2}}}+{\frac {d^{2}y}{du^{2}}}\left(4\,{\frac {du}{dx}}{\frac {d^{3}u}{dx^{3}}}+3\,\left({\frac {d^{2}u}{dx^{2}}}\right)^{2}\right)+{\frac {dy}{du}}{\frac {d^{4}u}{dx^{4}}}.\end{aligned}}} Salah satu bukti aturan rantai dimulai dengan definisi turunannya: ( f ∘ g ) ′ ( a )lim x → a f ( g ( x ) ) − f ( g ( a ) ) x − a .
{\frac {f(g(x))-f(g(a))}{x-a}}.} Assume for the moment that g ( x ) does not equal g ( a ) for any x near a .
{\frac {f(g(x))-f(g(a))}{x-a}}.} Asumsikan untuk saat itu g ( x ) tidak sama g ( a ) untuk apapun x dekat a. Maka persamaan sebelumnya sama dengan hasil kali dua faktor: lim x → a f ( g ( x ) )
{\frac {f(g(x))-f(g(a))}{g(x)-g(a)}}\cdot {\frac {g(x)-g(a)}{x-a}}.}
{\frac {f(g(x))-f(g(a))}{g(x)-g(a)}}\cdot {\frac {g(x)-g(a)}{x-a}}.} <!-+If
If g oscillates near a, then it might happen that no matter how close one gets to a, there is always an even closer x such that g(x)g(a).
g oscillates near a, then it might happen that no matter how close one gets to a, there is always an even closer x such that g ( x ) equals g ( a ) .
To work around this, introduce a function Q as follows: Q ( y )
Untuk menyiasatinya, perkenalkan sebuah fungsi Q sebagai berikut: Q ( y )
\displaystyle {\frac {f(y)-f(g(a))}{y-g(a)}},&y\neq g(a),\\f'(g(a)),&y=g(a).\end{cases}}} We will show that the difference quotient for f ∘ g is always equal to: Q ( g ( x ) ) ⋅ g ( x ) − g ( a ) x − a .
{\frac {f(y)-f(g(a))}{y-g(a)}},&y\neq g(a),\\f'(g(a)),&y=g(a).\end{cases}}} Kami akan menunjukkan bahwa hasil bagi perbedaan f ∘ g selalu sama dengan: Q ( g ( x ) ) ⋅ g ( x )
− f ( g ( a ) ) .
− g ( a ) x − a .
∑ i1 k ( d d x g i ( x ) )
D f n∏ k1 n [ D f k ∘ f ( k + 1 ) . .
D i f ( g 1 ( x ) , … , g k ( x ) ) .
− f ( g ( a ) ) g ( x ) − g ( a ) ⋅ g ( x ) − g ( a ) x − a .
f=u^{v}\ln u.} It follows that d d x ( g ( x ) h ( x ) )h ( x ) g ( x ) h ( x ) − 1 d d x g ( x ) + g ( x )
d d x ( f ( x ) g ( x ) )d d x ( f ( x ) ⋅ 1 g ( x ) )f ′ ( x ) ⋅ 1 g ( x ) + f ( x ) ⋅ d d x ( 1 g ( x ) ) .
{dx}}\left(g(x)^{h(x)}\right)=h(x)g(x)^{h(x)-1}{\frac {d}{dx}}g(x)+g(x)^{h(x)}\ln g(x)\,{\frac {d}{dx}}h(x).}
( f ∘ g ) ( x ) , maka d z d x | x( f ∘ g ) ′ ( x ) {dx}}\right|_{x}=(f\circ g)'(x)} dan d z d y | y ( x ) ⋅ d y d x | xf ′ ( y ( x ) )
From this perspective the chain rule therefore says: J f ∘ g ( a )J f ( g ( a ) ) J g ( a ) , (\mathbf {a} )
g dan h. Menerapkan aturan rantai dengan cara ini akan menghasilkan: ( f ∘ g ∘ h ) ′ ( a )
=J_{f}(g(\mathbf {a} ))
f ′ ( g ( x ) )
( g ′ ( a ) ) , J f ( g ( a ) )( f ′ ( g ( a ) ) ) .
{ f ( y ) − f ( g ( a ) ) y − g ( a ) , y ≠ g ( a ) , f ′ ( g ( a ) ) , yg ( a ) .