قاعدة السلسلة

في التحليل الرياضي، قاعدة السلسلة هي إحدى صيغ اشتقاق تركيب تابعين رياضيين:

F=f\circ g=f(g(x))

المشتق :

F'={\frac {dF}{dx}}=f'(g(x))\times g'(x).

بترميز لايبنتس:

{\frac {dy}{dx}}={\frac {dy}{du}}\times {\frac {du}{dx}}

More precisely, to indicate the point at each derivative is evaluated at, $\left.{\frac {dz}{dx}}\right|_{x}=\left.{\frac {dz}{dy}}\right|_{y(x)}\cdot \left.{\frac {dy}{dx}}\right|_{x}$ .

The versions of the chain rule in the Lagrange and the Leibniz notation are equivalent, in the sense that if $z=f(y)\!$ and $y=g(x)\!$ , so that $z=f(g(x))=(f\circ g)(x)$ , then

\left.{\frac {dz}{dx}}\right|_{x}=(f\circ g)'(x)

and

\left.{\frac {dz}{dy}}\right|_{y(x)}\cdot \left.{\frac {dy}{dx}}\right|_{x}=f'(y(x))g'(x)=f'(g(x))g'(x).

^[1]

In integration, the counterpart to the chain rule is the substitution rule.

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التاريخ

The chain rule seems to have first been used by Gottfried Wilhelm Leibniz. He used it to calculate the derivative of ${\sqrt {a+bz+cz^{2}}}$ as the composite of the square root function and the function $a+bz+cz^{2}\!$ . He first mentioned it in a 1676 memoir (with a sign error in the calculation). The common notation of chain rule is due to Leibniz.^[2] Guillaume de l'Hôpital used the chain rule implicitly in his 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.

بـُعـْد واحد

المثال الأول

Suppose that a skydiver jumps from an aircraft. Assume that $t$ seconds after his jump, his height above sea level in meters is given by $g (t) = 4000 - 4.9 t 2$ . One model for the atmospheric pressure at a height $h$ is $f (h) = 101325 e -0.0001 h$ . These two equations can be differentiated and combined in various ways to produce the following data:

$g'(t) = -9.8 t$ is the velocity of the skydiver at time $t$ .
$f'(h) = -10.1325 e -0.0001 h$ is the rate of change in atmospheric pressure with respect to height at the height $h$ and is proportional to the buoyant force on the skydiver at $h$ meters above sea level. (The true buoyant force depends on the volume of the skydiver.)
$(f \circ g)(t)$ is the atmospheric pressure the skydiver experiences $t$ seconds after his jump.
$(f \circ g)'(t)$ is the rate of change in atmospheric pressure with respect to time at $t$ seconds after the skydiver's jump, and is proportional to the buoyant force on the skydiver at $t$ seconds after his jump.

Here, the chain rule gives a method for computing $(f \circ g)'(t)$ in terms of $f'$ and $g'$ . While it is always possible to directly apply the definition of the derivative to compute the derivative of a composite function, this is usually very difficult. The utility of the chain rule is that it turns a complicated derivative into several easy derivatives.

The chain rule states that, under appropriate conditions,

(f\circ g)'(t)=f'(g(t))\cdot g'(t).

In this example, this equals

(f\circ g)'(t)={\big (}{\mathord {-}}10.1325e^{-0.0001(4000-4.9t^{2})}{\big )}\cdot {\big (}{\mathord {-}}9.8t{\big )}.

In the statement of the chain rule, $f$ and $g$ play slightly different roles because $f'$ is evaluated at $g(t)\!$ , whereas $g'$ is evaluated at $t$ . This is necessary to make the units work out correctly.

For example, suppose that we want to compute the rate of change in atmospheric pressure ten seconds after the skydiver jumps. This is $(f \circ g)'(10)$ and has units of pascals per second. The factor $g'(10)$ in the chain rule is the velocity of the skydiver ten seconds after his jump, and it is expressed in meters per second. $f'(g(10))\!$ is the change in pressure with respect to height at the height $g (10)$ and is expressed in pascals per meter. The product of $f'(g(10))\!$ and $g'(10)\!$ therefore has the correct units of pascals per second.

Here, notice that it is not possible to evaluate $f$ anywhere else. For instance, the 10 in the problem represents ten seconds, while the expression $f'(10)\!$ would represent the change in pressure at a height of ten meters, which is not what we wanted. Similarly, while $g'(10) = -98$ has a unit of meters per second, the expression $f'(g'(10))$ would represent the change in pressure at a height of −98 meters, which is again not what we wanted. However, $g (10)$ is 3020 meters above sea level, the height of the skydiver ten seconds after his jump, and this has the correct units for an input to $f$ .

Statement

The simplest form of the chain rule is for real-valued functions of one real variable. It states that if $g$ is a function that is differentiable at a point $c$ (i.e. the derivative $g'(c)$ exists) and $f$ is a function that is differentiable at $g (c)$ , then the composite function $f \circ g$ is differentiable at $c$ , and the derivative is^[3]

(f\circ g)'(c)=f'(g(c))\cdot g'(c).

The rule is sometimes abbreviated as

(f\circ g)'=(f'\circ g)\cdot g'.

If $y = f (u)$ and $u = g (x)$ , then this abbreviated form is written in Leibniz notation as:

{\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}.

^[1]

The points where the derivatives are evaluated may also be stated explicitly:

\left.{\frac {dy}{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 $f_{1},\ldots ,f_{n}\!$ with the composite function $f_{1}\circ (f_{2}\circ \cdots (f_{n-1}\circ f_{n}))\!$ , 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):

{\frac {df_{1}}{dx}}={\frac {df_{1}}{df_{2}}}{\frac {df_{2}}{df_{3}}}\cdots {\frac {df_{n}}{dx}}.

^[4]

المشتقات العليا

Faà di Bruno's formula generalizes the chain rule to higher derivatives. Assuming that $y = f (u)$ and $u = g (x)$ , then the first few derivatives are:

{\begin{aligned}{\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}}

البراهين

البرهان الأول

One proof of the chain rule begins with the definition of the derivative:

(f\circ g)'(a)=\lim _{x\to 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 $a$ . Then the previous expression is equal to the product of two factors:

\lim _{x\to a}{\frac {f(g(x))-f(g(a))}{g(x)-g(a)}}\cdot {\frac {g(x)-g(a)}{x-a}}.

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 $g(x)\!$ equals $g(a)\!$ . For example, this happens for $g (x) = x 2 sin(1 / x)$ near the point $a = 0$ . Whenever this happens, the above expression is undefined because it involves division by zero. To work around this, introduce a function $Q\!$ as follows:

Q(y)={\begin{cases}{\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 $f \circ g$ is always equal to:

Q(g(x))\cdot {\frac {g(x)-g(a)}{x-a}}.

Whenever $g (x)$ is not equal to $g (a)$ , this is clear because the factors of $g (x) - g (a)$ cancel. When $g (x)$ equals $g (a)$ , then the difference quotient for $f \circ g$ is zero because $f (g (x))$ equals $f (g (a))$ , and the above product is zero because it equals $f'(g (a))$ times zero. So the above product is always equal to the difference quotient, and to show that the derivative of $f \circ g$ at $a$ exists and to determine its value, we need only show that the limit as $x$ goes to $a$ of the above product exists and determine its value.

To do this, recall that the limit of a product exists if the limits of its factors exist. When this happens, the limit of the product of these two factors will equal the product of the limits of the factors. The two factors are $Q (g (x))$ and $(g (x) - g (a)) / (x - a)$ . The latter is the difference quotient for $g$ at $a$ , and because $g$ is differentiable at $a$ by assumption, its limit as $x$ tends to $a$ exists and equals $g'(a)$ .

As for $Q (g (x))$ , notice that $Q$ is defined wherever $f$ is. Furthermore, $f$ is differentiable at $g (a)$ by assumption, so $Q$ is continuous at $g (a)$ , by definition of the derivative. The function $g$ is continuous at $a$ because it is differentiable at $a$ , and therefore $Q \circ g$ is continuous at $a$ . So its limit as $x$ goes to $a$ exists and equals $Q (g (a))$ , which is $f'(g (a))$ .

This shows that the limits of both factors exist and that they equal $f'(g (a))$ and $g'(a)$ , respectively. Therefore, the derivative of $f \circ g$ at a exists and equals $f'(g (a))$ $g'(a)$ .^[4]