Course: 3.8 区别规则:链条规则

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区别规则:链条规则
The Chain Rule enables us to differentiate a composite function $\begin{align*}f \circ g\end{align*}$ . But why is there the need to have a special way to determine the derivative of a composite function? Intuitively, it is because the variation of the domain of $\begin{align*}f\end{align*}$ is now governed by the function $\begin{align*}g(x)\end{align*}$ rather than just by $\begin{align*}x\end{align*}$ , and $\begin{align*}g\end{align*}$ 's rate of change with respect to $\begin{align*}x\end{align*}$ should somehow be taken into account. Before proceeding, see if you find effect of $\begin{align*}g\end{align*}$ by comparing the derivative of $\begin{align*}f(x)=x^2\end{align*}$ with the derivative of $\begin{align*}f(x)=(5x)^2\end{align*}$ where $\begin{align*}g(x)=5x\end{align*}$ .
::链条规则允许我们区分复合函数 fg。但为什么需要有一个特殊的方法来确定复合函数的衍生物? 直观地说,这是因为 f 域的变异现在由函数 g(x) 而不是仅由 x 来调节, g 对 x 的变速率应该以某种方式得到考虑。在继续之前, 看看您是否通过比较 f(x) =x2 的衍生物和 f(x) = (5x) 2 的衍生物来发现 g 的效果, g (x) = 5x 。 g (x) = 5x 。

The Chain Rule
::链条规则

We want to derive a rule for the derivative of a composite function of the form $\begin{align*}f\circ g\end{align*}$ in terms of the of $\begin{align*}f\end{align*}$ and $\begin{align*}g\end{align*}$ . This rule would allow us to differentiate complicated functions in terms of known derivatives of simpler functions.
::我们希望从f.g.和f.g.形式的复合函数衍生出一项规则。这一规则将使我们能够在已知的较简单函数衍生物方面区分复杂的功能。

The rule that enables this is called the Chain Rule:
::允许这样做的规则称为“链条规则”:

If $\begin{align*}g\end{align*}$ is a differentiable function at $\begin{align*}x\end{align*}$ , and $\begin{align*}f\end{align*}$ is differentiable at $\begin{align*}g(x)\end{align*}$ , then the composition function $\begin{align*}f \circ g=f(g(x))\end{align*}$ is differentiable at $\begin{align*}x\end{align*}$ . The derivative of the composite function is:
::如果 g 是 x 的可区别函数, f 是 g( x) 的可区别函数, 那么组成函数 fg=f( g( x) 在 x 的可区别函数。复合函数的衍生物是:

$\begin{aligned} (f \circ g)^{'} (x) = f^{'} (g (x)) g^{'} (x) . \end{aligned}$
$\begin{align*}(f \circ g)^\prime (x)=f^\prime(g(x))g^\prime(x).\end{align*}$
:f*g) =(x) =(g(x)) g* (x)。

Or an equivalent statement:
::或同等声明:

If $\begin{align*}u=u(x)\end{align*}$ and $\begin{align*}f=f(u)\end{align*}$ , then $\begin{align*}\frac{d}{dx}[f(u)]=f^\prime (u) \frac{du}{dx}\end{align*}$
::如果 u= u( x) 和 f= f( u), 那么 ddx[ f( u)] = f_ (u) dudx

Or another equivalent statement:
::或另一种等同声明:

If $\begin{align*}y\end{align*}$ is a function of $\begin{align*}u\end{align*}$ , and $\begin{align*}u\end{align*}$ is a function of $\begin{align*}x\end{align*}$ , then
::如果 y 是 u 的函数, u 是 x 的函数, 那么

$\begin{aligned} \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \end{aligned}$
$\begin{align*}\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}\end{align*}$
::dydx=dydududx ddx dx ddx ddx ddx dx ddx ddx ddx

Apply the chain rule to find the derivative of $\begin{align*}f(x)=(2x^3-4x^2+5)^2\end{align*}$ .
::应用链规则查找 f( x) =( 2x3- 4x2+5) 的衍生物。 2

Using the chain rule, let $\begin{align*}u=2x^3-4x^2+5\end{align*}$ . Then
::使用链规则, 让 u= 2x3- 4x2+5 。

$\begin{aligned} \frac{d}{d x} [(2 x^{2} - 4 x^{2} + 5)^{2}] & = \frac{d}{d x} [u^{2}] \\ = 2 u \frac{d u}{d x} \\ = 2 (2 x^{3} - 4 x^{2} + 5) (6 x^{2} - 8 x) . \end{aligned}$
$\begin{align*}\frac{d}{dx}[(2x^2-4x^2+5)^2]&= \frac{d}{dx}[u^2] \\ &= 2 u \frac{du}{dx} \\ &= 2(2x^3-4x^2+5)(6x^2-8x).\end{align*}$
::dx[(2x2-4x2+5)2]=ddx[u2]=2ududx=2(2x3-4x2+5)(6x2-8x)

The problem above is one of the most common types of composite functions. It is a power function of the type
::上述问题是最常见的复合功能类型之一。

$\begin{align*}y=[u(x)]^n\end{align*}$ .
::y=[u(x)]n。

The rule for differentiating such functions is special case of the Chain Rule called the General Power Rule:
::区分这些职能的规则是称为 " 普遍权力规则 " 的链规则的特殊情形:

If $\begin{align*}y=[u(x)]^n\end{align*}$ , then $\begin{align*}\frac{dy}{dx}=n[u(x)]^{n-1} \frac{d}{dx}u(x)\end{align*}$
::如果 y= [u( x)]n, 那么 dydx=n [u( x)] n- 1dddxu( x)

Examples
::实例

Example 1
::例1

Earlier, you were asked if you find the effect of $\begin{align*}g\end{align*}$ on the derivative by comparing the derivative of $\begin{align*}f(x)=x^2\end{align*}$ with the derivative of $\begin{align*}f(x)=(5x)^2\end{align*}$ where $\begin{align*}g(x)=5x\end{align*}$ . The derivative of $\begin{align*}f(x)=x^2\end{align*}$ is $\begin{align*}f^\prime (x)=2x\end{align*}$ , and the derivative of $\begin{align*}f(x)=(5x)^2\end{align*}$ is $\begin{align*}f^\prime (x)=2(5x)5=2x \cdot 25\end{align*}$ . The effect of $\begin{align*}g\end{align*}$ in the composite function is to modify the rate of change of $\begin{align*}f(x)=x^2\end{align*}$ .
::早些时候,有人询问您是否通过将衍生物(x)=x2与 f(x) =(5x) 2 whereg(x)=5x的衍生物进行比较,发现衍生物中的g效应。衍生物(x)=x2 isf*(x)=2x,衍生物(x)=(5x)2)=(x)=(2x)=2(5x)5=2(x)25。混合函数的外引力效果是修改变化率(x)=x2。

Example 2
::例2

What is the slope of the tangent line to the function $\begin{align*}y=\sqrt{x^2-3x+2}\end{align*}$ that passes through point $\begin{align*}x=3\end{align*}$ ?
::通过点 x=3 的 y=x2 - 3x+2 函数的正切线斜度是多少?

We can write $\begin{align*}y=(x^2-3x+2)^{\frac{1}{2}}\end{align*}$ . This example illustrates the point that $\begin{align*}n\end{align*}$ can be any real number including fractions. Using the General Power Rule,
::我们可以写 Y = (x2- 3x+2) 12。这个示例可以说明 n 可以是包括分数在内的任何实际数字。使用一般权力规则,

$\begin{aligned} \frac{d y}{d x} & = \frac{1}{2} (x^{2} - 3 x + 2)^{\frac{1}{2} - 1} (2 x - 3) \\ = \frac{1}{2} (x^{2} - 3 x + 2)^{- \frac{1}{2}} (2 x - 3) \\ = \frac{(2 x - 3)}{2 \sqrt{x^{2} - 3 x + 2}} \end{aligned}$
$\begin{align*}\frac{dy}{dx}&= \frac{1}{2}(x^2-3x+2)^{\frac{1}{2}-1}(2x-3) \\ &= \frac{1}{2}(x^2-3x+2)^{-\frac{1}{2}}(2x-3) \\ &= \frac{(2x-3)}{2\sqrt{x^2-3x+2}}\end{align*}$
::uddx=12(x2-3x+2)12-1(2x-3)=12(x2-3x+2)--12(2x-3)=(2x-3)-2-3x+2

To find the slope of the tangent line, we simply substitute $\begin{align*}x=3\end{align*}$ into the derivative:
::为了找到正切线的斜坡, 我们只需将 x=3 替换为衍生物 :

$\begin{aligned} \frac{d y}{d x} |_{x = 3} = \frac{2 (3) - 3}{2 \sqrt{3^{2} - 3 (3) + 2}} = \frac{3}{2 \sqrt{2}} = \frac{3 \sqrt{2}}{4} . \end{aligned}$
$\begin{align*}\frac{dy}{dx} \bigg|_{x=3}=\frac{2(3)-3}{2\sqrt{3^2-3(3)+2}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}.\end{align*}$
::x=3=2(3)-3232-3(3)+2=322=324。

Example 3
::例3

Find $\begin{align*}\frac{dy}{dx}\end{align*}$ for $\begin{align*}y=\sin^3 x\end{align*}$ .
::查找 y=sin3x 的 dydx 。

The function can be written as $\begin{align*}y=[\sin x]^3\end{align*}$ . Thus
::函数可以以 y = [sin_x] 3 写入。

$\begin{aligned} \frac{d y}{d x} & = \frac{d}{d x} [\sin x]^{3} \\ = 3 [\sin x]^{2} [\cos x] \\ = 3 \sin x^{2} \cos x \end{aligned}$
$\begin{align*}\frac{dy}{dx}&= \frac{d}{dx}[\sin x]^3 \\ &= 3[\sin x]^2 [\cos x] \\ &= 3 \sin x^2 \cos x\end{align*}$
::dydx=ddx[sinx]3=3[sinx]2[cosx]=3sinx2cos*x

Example 4
::例4

Find $\begin{align*}\frac{dy}{dx}\end{align*}$ for $\begin{align*}y=[\cos (\pi x^2)]^3\end{align*}$ .
::查找 y= [cos(xx2) 的 dydx 。

This example will show the application of the chain rule multiple times because there are several functions embedded within each other.
::这个示例将多次显示链规则的应用,因为相互之间含有若干功能。

The function $\begin{align*}y\end{align*}$ can be written in the form
::y 函数可以用窗体写成

$\begin{align*}y=(u(w))^3\end{align*}$ , where
::y=(u(w))3, 其中

$\begin{aligned} u (w) & = \cos (w) \\ w (x) & = π x^{2} \end{aligned}$
$\begin{align*}u(w)&= \cos(w) \\ w(x)&= \pi x^2\end{align*}$
::u(w)=cos(w)w(x)x2

Here are the steps for the solution:
::以下是解决问题的步骤:

$\begin{aligned} \frac{d y}{d x} & = \frac{d}{d x} [u (w)^{3}] & \dots Use u and w substitutions . \\ = 3 \cdot u (w)^{2} \cdot \frac{d u}{d x} & \dots After using the General Power Rule \\ = 3 \cdot u (w)^{2} \cdot (\frac{d u}{d w} \cdot \frac{d w}{d x}) & \dots After using the Chain Rule for \frac{d u}{d x} \\ = 3 \cdot u (w)^{2} \cdot [- \sin (w) \cdot 2 π x] & \dots After evaluating \frac{d u}{d w} and \frac{d w}{d x} \\ = 3 [\cos (π x^{2})]^{2} \cdot (- \sin (π x^{2}) \cdot 2 π x) & \dots After substituting for u and w . \\ = - 6 π x [\cos (π x^{2})]^{2} \sin (π x^{2}) & \dots After simplification . \end{aligned}$
$\begin{align*}\frac{dy}{dx}&=\frac{d}{dx}[u(w)^3] \quad &&\ldots \text{Use} \ u \ \text{and} \ w \ \text{substitutions}. \\ &=3 \cdot u(w)^2 \cdot \frac{du}{dx} \quad &&\ldots \text{After using the General Power Rule} \\ &=3 \cdot u(w)^2 \cdot \left ( \frac{du}{dw} \cdot \frac{dw}{dx}\right ) \quad &&\ldots \text{After using the Chain Rule for} \frac{du}{dx} \\ &=3 \cdot u(w)^2 \cdot [-\sin (w) \cdot 2 \pi x] \quad &&\ldots \text{After evaluating} \ \frac{du}{dw} \ \text{and} \ \frac{dw}{dx} \\ &=3[\cos (\pi x^2)]^2 \cdot (-\sin (\pi x^2) \cdot 2 \pi x) \quad &&\ldots \text{After substituting for} \ u \ \text{and} \ w. \\ &=-6 \pi x[\cos (\pi x^2)]^2 \sin(\pi x^2) \quad &&\ldots \text{After simplification}.\end{align*}$
::在使用总电源规则=3(w)2(dudw)2(dudw)2(dudwüddx)...在使用链规则foldudx=3(w)2[-sin(w))2[-sin(w)22x]......在使用总电源规则=3(w)2(sin)2⁄2(sin)3[cos(x)2}2(x})2(sin(x2)2)2(sin)3]......后,在使用总电源规则=3(w)2(x2)2(x)2(xxxx)...之后,在使用电源规则[cos(x)2x]2(x)2(x)2(x)2(cos)之后,随后,在调换u和 w(x)6(x)6(x)2)。

Notice that we first used the General Power Rule and then used the Chain Rule, in the last step, we took the derivative of the argument.
::请注意,我们首先使用一般权力规则,然后使用链规则,在最后一步,我们采用了论点的衍生物。

Review
::回顾

For #1-11, find $\begin{align*}f^\prime (x)\end{align*}$ .
::#1-11,找找F_(x).

$\begin{align*}f(x)=(2x^2-3x)^{39}\end{align*}$
:xx) = (2x2-3x) 39

$\begin{align*}f(x)=\left ( x^3-\frac{5}{x^2}\right )^{-3}\end{align*}$
:xx)=(x3-5x2)-3

$\begin{align*}f(x)=\frac{1}{\sqrt{3x^2-6x+2}}\end{align*}$
:x) =13x2 - 6x+2

$\begin{align*}f(x)=\sin^3 x\end{align*}$
:xx) =sin3x

$\begin{align*}f(x)=\sin x^3\end{align*}$
:xx) =sinx3

$\begin{align*}f(x)=\sin^3 x^3\end{align*}$
:xx) =sin3x3

$\begin{align*}f(x)=\tan(4x^5)\end{align*}$
::f(x) = tan(4x5)

$\begin{align*}f(x)=\sqrt{4x- \sin^2 2x}\end{align*}$
:x)=4x-sin2=2x

$\begin{align*}f(x)=\frac{\sin x}{\cos(3x-2)}\end{align*}$
:xx) =sinxcos(3x-2)

$\begin{align*}f(x)=(5x+8)^3(x^3+7x)^{13}\end{align*}$
:xx) = (5x+8) 3(x3+7x) 13

$\begin{align*}f(x)=\left ( \frac{x-3}{2x-5}\right )^3\end{align*}$
:xx)=(x-32x-5)3

Find $\begin{align*}\frac{dy}{dx}\end{align*}$ for $\begin{align*}y=5 \cos(3x^2-1)\end{align*}$ .
::查找 y= 5cos @ (3x2- 1) 的 dydx 。

Find the derivative of $\begin{align*}\sqrt{x^3+x^5+89}\end{align*}$ .
::查找 x3+x5+89 的衍生物。

Find the derivative of $\begin{align*}\sin(\sin(\sin(x)))\end{align*}$ .
::查找罪的衍生物(sin(sin(x)))

By definition, any function composed with its inverse is just the identity: $\begin{align*}f(f^{-1}(x))=x\end{align*}$ . differentiate both sides of this equation and solve algebraically for the derivative of the inverse.
::根据其定义,与其反向构成的任何函数都只是身份:f(f-1(x)=x。区分此方程的两侧,并对反向衍生物进行代数解析。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

区别规则:链条规则

The Chain Rule ::链条规则

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

The Chain Rule
::链条规则

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)