Course: 3.10 隐性区别 | The Way To Learn

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隐隐性区分
The differentiation techniques up to this point have been applied to function definitions (equations) of the form $\begin{align*}y=f(x).\end{align*}$ Not all functions can be stated in this form, e.g., when products of $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ are involved. When they cannot, finding the derivative of $\begin{align*}y\end{align*}$ may require a different technique.
::至此为止的区分技术已应用于y=f(x)形式的功能定义(等式)。并非所有功能都可以这种形式表示,例如,当x和y的产品涉及时。在无法找到y的衍生物时,找到y的衍生物可能需要不同的技术。

Implicit Differentiation
::隐隐性区分

Consider the equation $\begin{align*}2xy=1\end{align*}$ .
::考虑2xy=1的方程。

We want to obtain the derivative $\begin{align*}\frac{dy}{dx}\end{align*}$ . One way to do this is to first solve for $\begin{align*}y\end{align*}$ , to produce an explicit function of $\begin{align*}x\end{align*}$ ,
::我们想要获得衍生元体。一个方法就是首先解决y, 产生x的明显功能,

$\begin{align*}y=\frac{1}{2x},\end{align*}$
::y=12x, y=12x, y=12x, y=12x, y=12x, y=12x, y=12x, y=12x, y=12x, y=12x,

and then take the derivative on both sides,
::然后,取出双方的派生,

$\begin{aligned} \frac{d y}{d x} & = \frac{d}{d x} [\frac{1}{2 x}] \\ = \frac{- 1}{2 x^{2}} . \end{aligned}$
$\begin{align} \frac{dy}{dx}&= \frac{d}{dx}\left [ \frac{1}{2x}\right ] \\ &= \frac{-1}{2x^2}.\\ \end{align}$
::dydx=ddx[12x]\\\\12x2。

However, there is another way of finding $\begin{align*}\frac{dy}{dx}\end{align*}$ that uses the original equation that is an implicit function of $\begin{align*}x\end{align*}$ . We can directly differentiate both sides.
::然而,还有另一种方法可以找到didx,它使用原方程式,即x的隐含功能。我们可以直接区分双方。

Find the derivative $\begin{align*}\frac{dy}{dx}\end{align*}$ of the equation $\begin{align*}2xy=1\end{align*}$ without transforming it to an explicit function of $\begin{align*}x\end{align*}$ . To do this, directly differentiate both sides:
::查找公式 2xy=1 的衍生元体, 不将其转换为 x 的明显函数。要做到这一点, 直接区分两边 :

$\begin{aligned} \frac{d}{d x} [2 x y] = \frac{d}{d x} [1] . \end{aligned}$
$\begin{align*}\frac{d}{dx}[2xy]=\frac{d}{dx}[1].\end{align*}$
::dx[2xy]=ddx[1]。

Using the Product Rule on the left-hand side,
::使用左手边的产品规则,

$\begin{aligned} y \frac{d}{d x} [2 x] + 2 x \frac{d}{d x} [y] & = 0 \\ y [2] + 2 x \frac{d y}{d x} & = 0. \end{aligned}$
$\begin{align*}y \frac{d}{dx}[2x]+2x \frac{d}{dx}[y]&= 0 \\ y[2]+2x \frac{dy}{dx}&= 0.\end{align*}$
::ydx[2x]+2xddx[y]=0y[2]+2xuddx=0。

Note that the is applied when taking the derivative of a term with $\begin{align*}y\end{align*}$ and a $\begin{align*}\frac{dy}{dx}\end{align*}$ is included for those terms. Solving for $\begin{align*}\frac{dy}{dx}\end{align*}$ ,
::请注意,这些术语中包含带 Y 和 dydx 的术语的衍生物时应用该术语。

$\begin{aligned} \frac{d y}{d x} = \frac{- 2 y}{2 x} = \frac{- y}{x} . \end{aligned}$
$\begin{align*}\frac{dy}{dx}=\frac{-2y}{2x}=\frac{-y}{x}.\end{align*}$
::2y2x x。

But since $\begin{align*}y=\frac{1}{2x}\end{align*}$ , substitution gives
::但因为 Y=12x, 替代给

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

This result agrees with the previous calculations. This second method of finding a derivative is called implicit differentiation. You may consider the process and say that the first method is easier and faster and there is no reason for the second method. That may be true for this example and some others, but consider the next problem.
::此结果与先前的计算结果一致。找到衍生物的第二种方法被称为隐含的区别。您可以考虑这个过程, 并说第一个方法更容易、更快, 没有理由采用第二个方法。这个例子和其他一些例子可能都是这样, 但是要考虑下一个问题。

Find $\begin{align*}\frac{dy}{dx}\end{align*}$ if $\begin{align*}3y^2- \cos y=x^3.\end{align*}$
::如果 3y2 - cosy=x3, 则查找 dydx 。

Good luck finding an explicit function representation of this equation. Let’s try implicit differentiation and see what happens.
::幸运地找到这个方程式的清晰功能代表。让我们尝试隐含的区别,看看会发生什么。

Differentiating both sides of the equation with respect to $\begin{align*}x\end{align*}$ and then solving for $\begin{align*}\frac{dy}{dx}\end{align*}$ ,
::将方程的两侧对齐 x 进行区分,然后解决 dydx,

$\begin{aligned} \frac{d}{d x} [3 y^{2} - \cos y] & = \frac{d}{d x} [x^{3}] \\ 3 \frac{d}{d x} [y^{2}] - \frac{d}{d x} [\cos y] & = 3 x^{2} \\ 3 (2 y \frac{d y}{d x}) - (- \sin y) \frac{d y}{d x} & = 3 x^{2} \\ 6 y \frac{d y}{d x} + \sin y \frac{d y}{d x} & = 3 x^{2} \\ [6 y + \sin y] \frac{d y}{d x} & = 3 x^{2} . \end{aligned}$
$\begin{align} \frac{d}{dx}[3y^2- \cos y]&= \frac{d}{dx}[x^3] \\ 3 \frac{d}{dx}[y^2]- \frac{d}{dx}[\cos y]&= 3x^2 \\ 3 \left(2y \frac{dy}{dx}\right)-(-\sin y) \frac{dy}{dx}&= 3x^2 \\ 6y \frac{dy}{dx}+ \sin y \frac{dy}{dx}&= 3x^2 \\ \left [ 6y+ \sin y \right ] \frac{dy}{dx}&= 3x^2.\\ \end{align}$
::ddx[3y2 -cosy] =ddx[x3] 3ddx[y2] -ddx[cosy] =3x23(2ydydx) - (-siny) dydx=3x26ydydx+sinydydx=3x2[6y+sinydx] dydx=3x2。

Solving for $\begin{align*}\frac{dy}{dx}\end{align*}$ , we finally obtain
::我们终于找到解毒剂的解毒方法了

$\begin{aligned} \frac{d y}{d x} = \frac{3 x^{2}}{6 y + \sin y} . \end{aligned}$
$\begin{align*}\frac{dy}{dx}=\frac{3x^2}{6y+ \sin y}.\end{align*}$
::dydx=3x26y+siny。

In this problem, implicit differentiation provided a workable path to a solution.
::在这个问题上,隐含的差别提供了可行的解决途径。

Implicit differentiation can be used to calculate the slope of the tangent line as the problem below shows.
::如下面的问题所示,可以使用隐含的差别来计算正切线的斜坡。

Find the equation of the tangent line that passes through the point (1, 2) on the graph of $\begin{align*}8y^3+x^2y-x=3\end{align*}$ .
::在 8y3+x2y-x=3 的图形中查找穿过点(1, 2)的正切线的方程。

The general approach to solving this problem is to:
::解决这一问题的一般办法是:

find $\begin{align*}\frac{dy}{dx}\end{align*}$ , then
::找到 dydx, 然后

substitute the point (1, 2) into the derivative to find the slope, and then
::将点(1, 2)替换为衍生物以找到斜度,然后

use the equation of the line (either the slope-intercept form or the point-intercept form) to find the equation of the tangent line.
::使用线条的方程( 斜度截面表或点截面表) 查找正切线的方程。

For step 1, finding an explicit function representation of the equation is not obvious. Using implicit differentiation, however, allows differentiation of both sides:
::对于第1步,在等式中找到明确的功能代表并不明显。

$\begin{aligned} \frac{d}{d x} [8 y^{3} + x^{2} y - x] & = \frac{d}{d x} [3] \\ 24 y^{2} \frac{d y}{d x} + [(x^{2}) (1) \frac{d y}{d x} + y (2 x)] - 1 & = 0 \\ 24 y^{2} \frac{d y}{d x} + x^{2} \frac{d y}{d x} + 2 x y - 1 & = 0 \\ [24 y^{2} + x^{2}] \frac{d y}{d x} & = 1 - 2 x y \\ \frac{d y}{d x} & = \frac{1 - 2 x y}{24 y^{2} + x^{2}} . \end{aligned}$
$\begin{align*}\frac{d}{dx}[8y^3+x^2y-x]&= \frac{d}{dx}[3] \\ 24y^2 \frac{dy}{dx}+ \left[(x^2)(1) \frac{dy}{dx}+ y(2x) \right]-1&= 0 \\ 24y^2 \frac{dy}{dx}+x^2 \frac{dy}{dx}+2xy-1&= 0 \\ \left [ 24y^2+x^2 \right ] \frac{dy}{dx}&= 1-2xy \\ \frac{dy}{dx}&= \frac{1-2xy}{24y^2+x^2}.\end{align*}$
::ddx[8y3]+x2y-x]=ddx[3,244y2dydx+[(x2)(1)uddx+y(2x)] -1=024y2dydx+x2uddx+2x2xy-1=0[24y2+x2]dydx=1-2xydydydx=1-2xy24y2+x2x2。

Now, substituting point (1, 2) into the derivative to find the slope,
::现在,用衍生物替代点(1,2) 找到斜坡

$\begin{aligned} \frac{d y}{d x} & = \frac{1 - 2 (1) (2)}{24 (2)^{2} + (1)^{2}} \\ = \frac{- 3}{97} . \end{aligned}$
$\begin{align*}\frac{dy}{dx}&= \frac{1-2(1)(2)}{24(2)^2+(1)^2} \\ &= \frac{-3}{97}.\end{align*}$
::dx=1-2(1)(2)24(2)2+(1)2397。

So the slope of the tangent line is $\begin{align*}\frac{-3}{97}\end{align*}$ , which is a small value. (What does this tell us about the orientation of the tangent line?)
::因此,正切线的斜坡是-397,这是一个很小的价值。 (这告诉我们正切线的方向是什么? )

Next we need to find the equation of the tangent line. The slope-intercept form is
::接下来我们需要找到正切线的方程。

$\begin{aligned} y = m x + b, \end{aligned}$
$\begin{align*}y=mx+b,\end{align*}$
::y=mx+b, y=mx+b, y=mx+b, y=mx+b, y=mx+b, y=mx+b, y=mx+b,

where $\begin{align*}m=\frac{-3}{97}\end{align*}$ and $\begin{align*}b\end{align*}$ is the $\begin{align*}y\end{align*}$ - intercept. To find $\begin{align*}b\end{align*}$ , simply substitute point (1, 2) into the line equation and solve:
::m397 和 b 是 Y 界面。要找到 b, 只需在直线方程式中找到替换点(1, 2) 并解析 :

$\begin{aligned} 2 & = (\frac{- 3}{97}) (1) + b \\ b & = \frac{197}{97} . \end{aligned}$
$\begin{align*}2&= \left ( \frac{-3}{97}\right )(1)+b \\ b&= \frac{197}{97}.\end{align*}$
::2=(-397)(1)+bb=1997。

Thus the equation of the tangent line is
::因此,正切线的方程式是

$\begin{aligned} y = \frac{- 3}{97} x + \frac{197}{97} . \end{aligned}$
$\begin{align*}y=\frac{-3}{97}x+ \frac{197}{97}.\end{align*}$
::y397x+1997.

Note that we could have used the equivalent point-slope form $\begin{align*}y-y_1=m(x-x_1)\end{align*}$ .
::请注意,我们本可以使用等效的点窗体y-y1=m(x-x1)。

To summarize, to find the derivative of an implicit function follow the following steps:
::简言之,为找到隐含功能的衍生物,可采取以下步骤:

Differentiate both sides with respect to $\begin{align*}x\end{align*}$
::区分双方对x的差别

Collect all $\begin{align*}\frac{dy}{dx}\end{align*}$ terms on the left hand side of the equation and place the other terms without $\begin{align*}\frac{dy}{dx}\end{align*}$ on the right hand side.
::收集方程式左侧的所有 dydx 条件, 并将其他条件放置在右手侧, 不 didx 条件。

Factor common $\begin{align*}\frac{dy}{dx}\end{align*}$ from all terms
::来自所有术语的公元数

Solve for $\begin{align*}\frac{dy}{dx}\end{align*}$ .
::解决子宫膜。

Note that the expression for the derivative $\begin{align*}\frac{dy}{dx}\end{align*}$ may involve BOTH $\begin{align*}x\end{align*}$ AND $\begin{align*}y\end{align*}$ .
::请注意,衍生面体的表达方式可能涉及两个 X 和 y 。

Examples
::实例

Example 1
::例1

Use implicit differentiation to find $\begin{align*}\frac{d^2y}{dx^2}\end{align*}$ if $\begin{align*}5x^2-4y^2=9\end{align*}$ .
::如果 5x2 - - 4y2=9, 则使用隐含差异来查找 d2ydx2 。

$\begin{aligned} \frac{d}{d x} [5 x^{2} - 4 y^{2}] & = \frac{d}{d x} [9] \\ 10 x - 8 y \frac{d y}{d x} & = 0. \end{aligned}$
$\begin{align*}\frac{d}{dx}[5x^2-4y^2]&= \frac{d}{dx}[9] \\ 10x-8y \frac{dy}{dx}&= 0.\end{align*}$
::ddx[5x2-4y2]=ddx[9]10x-8ydydx=0。

Solving for $\begin{align*}\frac{dy}{dx}\end{align*}$ ,
::正在解决 dydx,

$\begin{aligned} \frac{d y}{d x} = \frac{5 x}{4 y} . \end{aligned}$
$\begin{align*}\frac{dy}{dx}=\frac{5x}{4y}.\end{align*}$
::=5x4y

Differentiating both sides implicitly again (and using the quotient rule),
::再次暗含地区分双方(并使用商数规则),

$\begin{aligned} \frac{d^{2} y}{d x^{2}} & = \frac{(4 y) (5) - (5 x) (4 \frac{d y}{d x})}{(4 y)^{2}} \\ = \frac{20 y}{16 y^{2}} - \frac{20 x}{16 y^{2}} \frac{d y}{d x} \\ = \frac{5}{4 y} - \frac{5 x}{4 y} \frac{d y}{d x} . \end{aligned}$
$\begin{align*}\frac{d^2y}{dx^2}&= \frac{(4y)(5)-(5x)\left ( 4\frac{dy}{dx}\right )}{(4y)^2} \\ &=\frac{20y}{16y^2}- \frac{20x}{16y^2} \frac{dy}{dx}\\ &= \frac{5}{4y}- \frac{5x}{4y} \frac{dy}{dx}.\end{align*}$
::d2ydx2=(4y)(5)-(5x)(4uddx)(4y)2=20y16y2-20x16y2udidx=54y-5x4ydx)。

But since $\begin{align*}\frac{dy}{dx}=\frac{5x}{4y}\end{align*}$ , we substitute it into the second derivative:
::但既然Didx=5x4y, 我们把它换成第二个衍生物:

$\begin{aligned} \frac{d^{2} y}{d x^{2}} & = \frac{5}{4 y} - \frac{5 x}{4 y} \cdot \frac{5 x}{4 y} \\ \frac{d^{2} y}{d x^{2}} & = \frac{5}{4 y} - \frac{25 x^{2}}{16 y^{2}} . \end{aligned}$
$\begin{align*}\frac{d^2y}{dx^2}&= \frac{5}{4y}- \frac{5x}{4y} \cdot \frac{5x}{4y} \\ \frac{d^2y}{dx^2}&= \frac{5}{4y}- \frac{25x^2}{16y^2}.\end{align*}$
::d2ydx2 = 54y-5x4y_5x4y_5x4yd2yd2ydx2=54y-25x216y2。

This is the second derivative of $\begin{align*}y\end{align*}$ .
::这是y的第二个衍生物

Example 2
::例2

Use implicit differentiation to find $\begin{align*}\frac{d^2y}{dx^2} \Big|_{(x,y)=(2,3)}\end{align*}$ .
::使用隐含的区别来查找 d2ydx2 \\\ (x,y) = (2, 3) 。

The next step is to find: $\begin{align*}\frac{d^2y}{dx^2} \Big|_{(x,y)=(2,3)}\end{align*}$ .
::下一步是找到: d2ydx2(x,y) = (2,3) 。

$\begin{aligned} \frac{d^{2} y}{d x^{2}} |_{(2, 3)} & = \frac{5}{4 (2)} - \frac{25 (2)^{2}}{16 (3)^{2}} \\ = - \frac{5}{72} . \end{aligned}$
$\begin{align*}\frac{d^2 y}{dx^2} \bigg |_{(2,3)}&= \frac{5}{4(2)}- \frac{25(2)^2}{16(3)^2} \\ &= - \frac{5}{72}.\end{align*}$
::d2ydx2(2,3)=54(2)-25(2)(2)216(3)2572。

Example 3
::例3

What does the second derivative represent?
::第二个衍生物代表什么?

Since the first derivative of a function represents the rate of change of the function $\begin{align*}y=f(x)\end{align*}$ with respect to $\begin{align*}x\end{align*}$ , the second derivative represents the rate of change of the function. For example, in kinematics (the study of motion), the speed of an object $\begin{align*}(y^\prime)\end{align*}$ signifies the change of position with respect to time but acceleration $\begin{align*}(y^{\prime\prime})\end{align*}$ signifies the rate of change of the speed with respect to time.
::由于函数的第一个衍生物是函数y=f(x)相对于x的变动率,第二个衍生物代表函数的变动率。例如,在运动运动学(运动研究)中,对象的速度(y`)表示时间位置的变动,而加速表示时间速度的变动率。

Review
::回顾

For #1-6, find $\begin{align*}\frac{dy}{dx}\end{align*}$ by implicit differentiation.
::对于#1-6,通过隐含的区别来找到。

$\begin{align*}x^2+y^2=500\end{align*}$
::x2+y2=500

$\begin{align*}x^2y+3xy-2=1\end{align*}$
::x2y+3xy-2=1

$\begin{align*}\frac{1}{x}+ \frac{1}{y}=\frac{1}{2}\end{align*}$
::1x+1y=12

$\begin{align*}\sqrt{x}-\sqrt{y}=\sqrt{3}\end{align*}$
::x- y=3

$\begin{align*}\sin(25xy^2)=x\end{align*}$
:25xy2)=x

$\begin{align*}\tan^3(x^2-y^2)=\tan\left ( \frac{\pi}{4}\right )\end{align*}$
::tan_( % 4) (x2- y2) =tan_ (% 4)

For #7-8, use implicit differentiation to find the slope of the tangent line to the given curve at the specified point.
::对于 # 7-8, 使用隐含的区分法在指定点找到正切线与给定曲线的斜坡。

$\begin{align*}x^2y-y^2x=-1\end{align*}$ at (1,1)
::x2y-y2x1 时(1,1)

$\begin{align*}\sin(xy)=y\end{align*}$ at $\begin{align*}(\pi,1)\end{align*}$
::sin(xy)=y 时 (..., 1)

Find $\begin{align*}y^{\prime\prime}\end{align*}$ by implicit differentiation for $\begin{align*}x^3y^3=5\end{align*}$ .
::通过 x3y3=5 的隐含区别查找 y 。

Use implicit differentiation to show that the tangent line to the curve $\begin{align*}y^2=kx\end{align*}$ at $\begin{align*}(x_0,y_0)\end{align*}$ is given by $\begin{align*}y_0y=\frac{1}{2}k(x+x_0)\end{align*}$ , where $\begin{align*}k\end{align*}$ is a constant.
::使用隐含差异来显示( x0, y0) 时曲线 y2=kx 的正切线由 y0y= 12k( x+x0) 给出, k 是常数。

Find %2By%20%5Csin(x))"> $\begin{align*}\frac{d}{dx}(x \sin(y)+y \sin(x))\end{align*}$ .
::查找 ddx(xsin+ysin(x)) 。

Find $\begin{align*}y^\prime\end{align*}$ if $\begin{align*}x^2+xy+y^2=10\end{align*}$ .
::如果 x2+xy+y2=10,请查找 y`。

Find the formula for the line tangent to the curve $\begin{align*}y^3+2xy^2-x=2\end{align*}$ at the point (1, 1).
::在点(1, 1) 找到与曲线 y3+2xy2 - x=2 的正切线的公式。

Find $\begin{align*}\frac{d}{dx} \left ( \sqrt{xy} \right )\end{align*}$
::查找 ddx( xy)

%3Dx"> $\begin{align*}y^2+\sin(y)=x\end{align*}$ . Find $\begin{align*}\frac{d^2y}{dx^2}\end{align*}$ in terms of $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ .
::y2+sin=x. 查找 x 和 y 的 d2ydx2 。

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

隐隐性区分

Implicit Differentiation ::隐隐性区分

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Implicit Differentiation
::隐隐性区分

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)