微积分 | dy / dx 不是分数吗？

注：本文为 “微积分 | dy / dx” 相关讨论辨析合辑。
英文引文，机翻未校。
如有内容异常，请看原文。

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Is $\frac{dy}{dx}$ a fraction? Yes and no …

$\frac{dy}{dx}$ 是分数吗？既是也不是……

August 17, 2020 / By Dave Peterson

A question from 2004, the first of the two I directed Zanzabar to, looked more closely at the $\frac{dy}{dx}$ notation:
2004 年的一个问题（我给赞扎巴尔推荐的两个问题中的第一个）更深入地探讨了 $\frac{dy}{dx}$ 符号：

Can $\frac{dy}{dx}$ Be Treated as a Fraction?
$\frac{dy}{dx}$ 可以被当作分数对待吗？

When I learned about derivatives, I learned that $\frac{dy}{dx}$ was a notation that implied “derivative of $y$ with respect to $x$.” I understood that. But I am confused about whether or not the notation $\frac{dy}{dx}$ can be treated as a fraction, giving individual meanings to $dy$ and $dx$.
当我学习导数时，我知道 $\frac{dy}{dx}$ 是一个符号，表示“$y$ 对 $x$ 的导数”。这一点我理解。但我困惑的是，$\frac{dy}{dx}$ 这种符号是否可以被当作分数对待，从而赋予 $dy$ 和 $dx$ 各自的含义。

For example, in integration by substitution:
例如，在换元积分法中：

integrate ($\sin (3x+5)dx$)
求 $\int \sin (3x+5)dx$

$u=3x+5$

$\frac{du}{dx}=3$

$dx=\frac{du}{3}$

That is the part that confuses me. How can the $dx$ be solved for? What exactly is “$dx$”?
这正是让我困惑的地方。怎么能解出 $dx$ 呢？“$dx$”到底是什么？

I understand that $\frac{dy}{dx}$ is a limit, and that it is a slope. But the idea of it being simply a notation doesn’t help me understand how you can multiply out the bottom. Any help would be appreciated…
我知道 $\frac{dy}{dx}$ 是一个极限，也是一个斜率。但仅仅将其视为一种符号，无法帮助我理解为什么可以将分母乘过去。希望能得到您的帮助……

This separate use of $dx$ and $dy$ is particularly common in integration, which is the inverse of differentiation, but is also used in other ways. Is it really legal to break $\frac{dy}{dx}$ apart like this?
$dx$ 和 $dy$ 的这种单独使用在积分（微分的逆运算）中尤为常见，但也用于其他方面。像这样将 $\frac{dy}{dx}$ 拆开真的合理吗？

This topic was previously discussed in less depth, in the post Why Do People Treat $\frac{dy}{dx}$ as a Fraction?, where I quoted an unarchived answer from 2015, and gave links to some of the pages we’re looking at here.
这个主题之前在《为什么人们把 $\frac{dy}{dx}$ 当作分数对待？》一文中有过较浅层次的讨论，在那篇文章中，我引用了 2015 年一个未存档的答案，并给出了我们在这里所看的一些页面的链接。

No …

不是……

Doctor Vogler answered:
沃格勒博士回答：

Hi Amit,
你好，阿米特，

Thanks for writing to Dr Math. The easy answer to your question is that your definition for $\frac{dy}{dx}$ is correct; it means the derivative of $y$ with respect to $x$, and $dy$ and $dx$ are meaningless when written alone, so that
感谢你给数学博士写信。你的问题的简单答案是，你对 $\frac{dy}{dx}$ 的定义是正确的；它表示 $y$ 对 $x$ 的导数，而 $dy$ 和 $dx$ 单独写出来是没有意义的，所以

$dx=\frac{du}{3}$

is not a meaningful expression but should be written
不是一个有意义的表达式，而应该写成

$\frac{dx}{du}=\frac{1}{3}$.

This follows the formal definition of $\frac{dy}{dx}$, which represents a single operator on a function, not an actual fraction. But certain things you can do with a derivative look an awful lot like what we can do with fractions:
这符合 $\frac{dy}{dx}$ 的正式定义，它表示对函数的一个单一算子，而不是一个实际的分数。但对导数进行的某些操作，看起来非常像我们对分数进行的操作：

And when certain nice things happen that look like fractions, such as:
当某些看起来像分数的奇妙现象出现时，例如：

$$\frac{dy}{dz}=\frac{1}{\frac{dz}{dy}}$$

and
以及

$$\frac{dz}{dx}=\frac{dz}{dy}\cdot \frac{dy}{dx}$$

then this is actually just the Chain Rule at work. And the reason that
那么这实际上只是链式法则在起作用。而

$\int f(g(x))g^{\prime }(x)dx=\int f(u)du$

is not that
并不是因为

$u=g(x)$

implies
意味着

$du=g^{\prime }(x)dx$

but rather the Chain Rule again.
而是再次因为链式法则。

The chain rule, and the related concept of substitution in an integral “just happen” to look like you can juggle $dx$ and $dy$ and $du$, called differentials, as if they were numbers themselves. And the rules for substitution in an integral are proved by the chain rule, not by treating differentials as real entities.
链式法则以及积分中相关的换元概念，“恰好”看起来像你可以像处理数字一样摆弄 $dx$、$dy$ 和 $du$（称为微分）。积分中的换元法则是由链式法则证明的，而不是通过将微分视为真实实体来证明的。

… But, yes, in fact

……但实际上，是的

All of that is true, except that I should qualify the “not a meaningful expression.” You see, something is only meaningless until somebody gives it a formal meaning. Then you hope that the meaning they gave it has useful properties (such as, that it relates to derivatives…). In fact, this has been done, and there is a good deal of mathematics that has gone into the theory of differentials, and it fits into integrals, and putting the differential “$dx$” at the end of every integral also makes sense according to this theory, and so on. One math doctor alluded to some of this on Differentials.
所有这些都是事实，但我应该对“不是一个有意义的表达式”进行限定。你看，某事物只有在有人赋予它正式意义之前才是无意义的。然后你会希望他们赋予的意义具有有用的性质（例如，与导数相关……）。事实上，人们已经做到了这一点，并且有大量的数学知识投入到了微分理论中，微分理论与积分相契合，根据这一理论，在每个积分的末尾加上微分“$dx$”也是有意义的，等等。

You can also get books that discuss this in more detail. But the fact is that most people who use calculus don’t really need all of the theory of differentials, and the Chain Rule indeed suffices to verify most facts that you would get from treating $\frac{dy}{dx}$ as a fraction. The reason you can treat it as a fraction is that $\frac{dy}{dx}$ is the limit of a fraction, and so most of the operations you would do to the fraction you can do before you take the limit. In other words, before the limit is taken, it is a fraction, so you can treat it as one. But then you take the limit and it becomes a derivative.
你也可以找到更详细讨论这个问题的书籍。但事实是，大多数使用微积分的人并不真正需要全部的微分理论，链式法则确实足以验证大多数将 $\frac{dy}{dx}$ 当作分数对待时得到的结论。你可以将其当作分数对待的原因是，$\frac{dy}{dx}$ 是一个分数的极限，因此你对分数进行的大多数操作，都可以在取极限之前进行。换句话说，在取极限之前，它就是一个分数，所以你可以将其当作分数对待。但当你取极限后，它就变成了导数。

We can say that the derivative “inherits” some (not all) of the behavior of fractions from the fraction (difference quotient) that is at the heart of its definition. That fraction is the motivation for the notation, and the notation makes it easy to remember things like the chain rule, but you have to remember that in spite of all this, it isn’t really a fraction, and you can only treat it as a fraction where there is a theorem that says you can.
我们可以说，导数从其定义核心的分数（差商）中“继承”了一些（并非全部）分数的性质。那个分数是这种符号的设计动机，而且这种符号让人们很容易记住像链式法则这样的内容，但你必须记住，尽管如此，它并不真的是一个分数，只有在有定理表明可以将其当作分数对待的情况下，你才能这样做。

One place where we actually write differentials on their own, besides integration, is in estimation, which ties in with the theory of differentials:
除了积分之外，我们实际上单独使用微分的一个地方是在估计中，这与微分理论相关：

Finally, there is also the theory of estimating with derivatives, where I always say to think of
最后，还有用导数进行估计的理论，在这个理论中，我总是说可以这样想：

$dx=$ change in $x$

$dy=$ change in $y$

$x=$ unchanged value of $x$

$y=$ unchanged value of $y$

For example, to estimate $(1.98{)}^6$, we use
例如，要估计 $(1.98{)}^6$，我们使用

$y=x^6$

$dy=6x^{5}dx$

$x=2$

$dx=-0.02$

(so that $x+dx=1.98$), and therefore
（这样 $x+dx=1.98$），因此

$y=2^6=64$

$dy=6x^{5}dx=6\cdot 32\cdot (-0.02)=-3.84$

which implies that
这意味着

$y+dy=64-3.84=60.16$,

which is, in fact, a pretty close approximation to $(1.98{)}^6$. And this is essentially treating $\frac{dy}{dx}$ as a fraction before we’ve taken the limit, since $dx$ doesn’t go all the way to zero but only to $-0.02$.
实际上，这与 $(1.98{)}^6$ 的值非常接近。这本质上是在取极限之前将 $\frac{dy}{dx}$ 当作分数对待，因为 $dx$ 并没有完全趋近于零，而只是到 $-0.02$。

Does this help you to understand how differentials are a fraction in some sense but not in others? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions.
这是否帮助你理解了微分在某种意义上是分数，而在其他意义上不是分数呢？如果你对此有任何疑问或需要更多帮助，请回信告诉我你目前的理解情况，我会尽力提供进一步的建议。

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Is this a correct/good way to think interpret differentials for the beginning calculus student?

对于初学微积分的学生来说，这种理解和解释微分的方式是否正确/恰当？

asked Nov 27, 2015 at 10:26，Ovi

I was reading the answers to this question, and I came across the following answer which seems intuitive, but too good to be true:

我在阅读这个问题（上文）的答案时，遇到了以下这个答案，它看起来很直观，但似乎好得令人难以置信：

Typically, the $\frac{dy}{dx}$ notation is used to denote the derivative, which is defined as the limit we all know and love (see Arturo Magidin’s answer). However, when working with differentials, one can interpret $\frac{dy}{dx}$ as a genuine ratio of two fixed quantities.
 
通常，$\frac{dy}{dx}$ 符号用于表示导数，其定义是我们熟知且常用的极限（见 Arturo Magidin 的回答）。然而，在处理微分时，人们可以将 $\frac{dy}{dx}$ 理解为两个固定量的真正比率。
 
Draw a graph of some smooth function $f$ and its tangent line at $x=a$. Starting from the point $(a,f(a))$, move $dx$ units rightalong the tangent line(not along the graph of $f$). Let $dy$ be the corresponding change in $y$.
 
画出某个光滑函数 $f$ 的图像及其在 $x=a$ 处的切线。从点 $(a,f(a))$ 开始，沿着切线（而非 $f$ 的图像）向右移动 $dx$ 个单位。令 $dy$ 为相应的 $y$ 的变化量。
 
So, we moved $dx$ units right, $dy$ units up, and stayed on the tangent line. Therefore the slope of the tangent line is exactly $\frac{dy}{dx}$. However, the slope of the tangent at $x=a$ is also given by $f^{\prime }(a)$, hence the equation $\frac{dy}{dx}=f^{\prime }(a)$ holds when $dy$ and $dx$ are interpreted as fixed, finite changes in the two variables $x$ and $y$.

也就是说，我们向右移动了 $dx$ 个单位，向上移动了 $dy$ 个单位，且始终在切线上。因此，切线的斜率恰好是 $\frac{dy}{dx}$。然而，$x=a$ 处切线的斜率也由 $f^{\prime }(a)$ 给出，因此当 $dy$ 和 $dx$ 被解释为变量 $x$ 和 $y$ 的固定、有限变化量时，等式 $\frac{dy}{dx}=f^{\prime }(a)$ 成立。

• 微分的一种直观理解：$\frac{dy}{dx}$ 可视为函数在 $x=a$ 处切线上的斜率，其中 $dx$ 是沿切线向右的水平变化量，$dy$ 是对应的垂直变化量，且 $\frac{dy}{dx}=f^{\prime }(a)$。
• 导数的符号 $\frac{dy}{dx}$ 在微分语境下可被直观解释为切线斜率的比率，但本质上导数的定义基于极限。

Answer by user195934

answered Jan 10, 2016 at 21:53

The conclusion is right, but you should not understand $\frac{dy}{dx}$ that way. When you do what you have done it is written $\frac{\Delta y}{\Delta x}$.

结论是对的，但你不应该那样理解 $\frac{dy}{dx}$。当你做你所描述的操作时，它应该被写成 $\frac{\Delta y}{\Delta x}$。

If you do what you have explained and take the fixed values, observe that you can get closer to $a$ on the tangent and do the same again.

如果你按照你所解释的那样取固定值，会发现你可以在切线上离 $a$ 更近的地方重复同样的操作。

A derivative of a sufficiently nice function is saying that no matter how close you get to $a$ using your tangent principle, the result is going to be the same. In that sense you are right, you can take any fixed value on the tangent, but fixing something islessgeneral than saying no matter what fixed value on the tangent you take.

对于 “足够好” 的函数，其导数的意义是：无论你用切线原理离 $a$ 多近，结果都是一样的。从这个意义上说，你是对的 —— 你可以在切线上取任何固定值，但 “固定某个值” 不如 “无论在切线上取什么固定值，结果都相同” 更具一般性。

Observe as well that the way you construct a tangent is not something that is logically above the definition of the derivative so you could say: we know how to construct a tangent and then we can argue about the consequences. In general, drawing a tangent and finding first derivative are equivalent.

还要注意，构造切线的方法在逻辑上并不高于导数的定义 —— 可以说，我们知道如何构造切线，然后才能讨论其结果。一般来说，画切线和求一阶导数是等价的。

When a function is sufficiently nice all things are clearer, but you must define differentiability so that it is applicable to a wider range of problems.

当函数 “足够好” 时，一切都更清晰，但必须定义可微性（differentiability），使其能应用于更广泛的问题。

You need to notice that turning $\frac{\Delta y}{\Delta x}$ into $\frac{dy}{dx}$ and approaching one and the same value is inthe coreof the definition of having a derivative.

你需要注意，将 $\frac{\Delta y}{\Delta x}$ 转化为 $\frac{dy}{dx}$ 并趋近于同一个值，这是导数定义的核心。

Believe or not, the way you have defined a derivative is applicable in another theory: the theory of chaos, since for many chaotic curves you cannot draw a tangent. Instead you take two close points find the distance and calculate $\frac{\Delta y}{\Delta x}$. In many cases you get a fixed value as you approach $\frac{dy}{dx}$, although it is not possible to draw a tangent in the classical sense. Even when you cannot get a fixed value you make some averaging and still get something useful.

信不信由你，你所定义导数的方式可应用于另一个理论 —— 混沌理论（theory of chaos）：因为对于许多混沌曲线，你无法画出切线。相反，你取两个接近的点，计算它们的距离并算出 $\frac{\Delta y}{\Delta x}$。在很多情况下，当你趋近于 $\frac{dy}{dx}$ 时，会得到一个固定值，尽管无法用经典意义上的方法画出切线。即使得不到固定值，你也可以通过某种平均得到有用的结果。

Basically $\frac{dy}{dx}={\lim }_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$ and that is the way you should understand it.

本质上，$\frac{dy}{dx}={\lim }_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$，这才是你应该理解它的方式。

But yes, you can find a derivative the way you have described for many nice behaving functions.

但确实，对于许多性质良好的函数，你可以用你所描述的方式求导数。

• $\frac{dy}{dx}$ 的本质是极限：$\frac{dy}{dx}={\lim }_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$，而非单纯的固定量比率（$\frac{\Delta y}{\Delta x}$ 是有限变化量的比率）。
• 导数与切线的关系：对于 “足够好” 的函数，画切线与求一阶导数等价；导数的定义确保了 “无论在切线上取多近的点，结果都一致”。
• 可微性的定义需具备一般性，以适用于更广泛的函数（包括非 “性质良好” 的函数）。
• 导数在混沌理论中的应用：通过计算接近点的 $\frac{\Delta y}{\Delta x}$ 趋近于导数，即使无法画出经典切线。

Answer by user137731

answered Jan 10, 2016 at 21:16

It’s my opinion that beginning calculus students shouldn’t be thinking about differentials at all. You should just take $\frac{dy}{dx}$ as a notation for the derivative. Trying to think of $dy$ and $dx$ as independent objects is just going to confuse you at this level – assuming your course doesn’t take Robinson’s approach.

我的观点是，初学微积分的学生根本不需要纠结微分。你只需把 $\frac{dy}{dx}$ 当作导数的符号即可。在这个阶段，试图将 $dy$ 和 $dx$ 视为独立的对象只会让你困惑 —— 除非你的课程采用罗宾逊的非标准分析方法。

I realize that you’ll probably have to go through a section on differentials in your calculus class but just reinterpret the questions from that section as questions about the tangent line to $y=y(x)$.

我知道你在微积分课上可能不得不学微分的内容，但只需把这部分的问题重新理解为 “关于 $y=y(x)$ 的切线” 的问题即可。

Here’s an example problem:

举个例子：

Use differentials to approximate the value of $\sqrt{3.98}$.

用微分近似计算 $\sqrt{3.98}$ 的值。

Seeing this you should immediately mentally convert the word “differentials” to “tangent line approximations”. The line tangent to a differentiable function $f$ at $a$ is given by $T(a+h)=f(a)+f^{\prime }(a)h$.

看到这个问题时，你应该立即在脑海中把 “微分” 转换成 “切线近似”。可微函数 $f$ 在 $a$ 处的切线方程为 $T(a+h)=f(a)+f^{\prime }(a)h$。

Your professor will probably expect you to use differential notation – unfortunately – so when you actually write this on your paper use the notation $\Delta y=y(x+dx)-y(x)\approx dy(x,dx)=y^{\prime }(x)dx$ but just keep in mind that this is only a change of notation – you’re still just doing a tangent line approximation problem.

不幸的是，你的教授可能希望你用微分符号，所以当你在纸上写的时候，用符号 $\Delta y=y(x+dx)-y(x)\approx dy(x,dx)=y^{\prime }(x)dx$，但要记住，这只是符号的转换 —— 你仍然在做切线近似问题。

Then the way to solve the above problem is like so (the blue is the part you should write on your homework, the red is the part you should be using on your scratch paper when actually working out the problem):

解决上述问题的方法如下（蓝色部分是你应该写在作业上的内容，红色部分是你在草稿纸上实际计算时应该用的内容）：

$y(x)+dy(x,dx)=T((4)+(-0.02))=\sqrt{4}+{(\sqrt{x}{)}^{\prime }|}_{x=4}(-0.02)=2+\frac{-0.02}{2\sqrt{4}}=1.995$

Which does in fact approximate $\sqrt{3.98}=\mathrm{1.994993...}$ pretty well.

这实际上很好地近似了 $\sqrt{3.98}=\mathrm{1.994993...}$。

• 对初学者的建议：暂时无需深入理解微分的本质，可将微分符号（如 $dy$、$dx$）视为切线近似的符号转换。
• 微分与切线近似的关系：$\Delta y\approx dy=y^{\prime }(x)dx$ 本质上是函数在 $x$ 处的切线近似，即 $T(x+dx)=y(x)+y^{\prime }(x)dx$。
• 微分的应用：通过切线近似计算函数值（如 $\sqrt{3.98}\approx 1.995$），核心是利用切线方程 $T(a+h)=f(a)+f^{\prime }(a)h$。

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Is This a Correct/Good Way to Think Interpret Differentials for the Beginning Calculus Student?

对于初学微积分的学生来说，这种理解和解释微分的方式是否正确/恰当？

Intuitive Understanding of Differentials

微分的直观理解

Typically, the $\frac{dy}{dx}$ notation is used to denote the derivative, which is defined as the limit we all know and love (see Arturo Magidin’s answer). However, when working with differentials, one can interpret $\frac{dy}{dx}$ as a genuine ratio of two fixed quantities.
通常，$\frac{dy}{dx}$ 符号用于表示导数，其定义是我们熟知且常用的极限（见 Arturo Magidin 的回答）。然而，在处理微分时，人们可以将 $\frac{dy}{dx}$ 理解为两个固定量的真正比率。
 
Draw a graph of some smooth function $f$ and its tangent line at $x=a$. Starting from the point $(a,f(a))$, move $dx$ units right along the tangent line (not along the graph of $f$). Let $dy$ be the corresponding change in $y$.
画出某个光滑函数 $f$ 的图像及其在 $x=a$ 处的切线。从点 $(a,f(a))$ 开始，沿着切线（而非 $f$ 的图像）向右移动 $dx$ 个单位。令 $dy$ 为相应的 $y$ 的变化量。
 
So, we moved $dx$ units right, $dy$ units up, and stayed on the tangent line. Therefore the slope of the tangent line is exactly $\frac{dy}{dx}$. However, the slope of the tangent at $x=a$ is also given by $f^{\prime }(a)$, hence the equation $\frac{dy}{dx}=f^{\prime }(a)$ holds when $dy$ and $dx$ are interpreted as fixed, finite changes in the two variables $x$ and $y$.
也就是说，我们向右移动了 $dx$ 个单位，向上移动了 $dy$ 个单位，且始终在切线上。因此，切线的斜率恰好是 $\frac{dy}{dx}$。然而，$x=a$ 处切线的斜率也由 $f^{\prime }(a)$ 给出，因此当 $dy$ 和 $dx$ 被解释为变量 $x$ 和 $y$ 的固定、有限变化量时，等式 $\frac{dy}{dx}=f^{\prime }(a)$ 成立。

Critique of the Intuitive Approach

对直观方法的批判

The conclusion is right, but you should not understand $\frac{dy}{dx}$ that way. When you do what you have done it is written $\frac{\Delta y}{\Delta x}$.
结论是对的，但你不应该那样理解 $\frac{dy}{dx}$。当你做你所描述的操作时，它应该被写成 $\frac{\Delta y}{\Delta x}$。

If you do what you have explained and take the fixed values, observe that you can get closer to $a$ on the tangent and do the same again.
如果你按照你所解释的那样取固定值，会发现你可以在切线上离 $a$ 更近的地方重复同样的操作。

A derivative of a sufficiently nice function is saying that no matter how close you get to $a$ using your tangent principle, the result is going to be the same. In that sense you are right, you can take any fixed value on the tangent, but fixing something is less general than saying no matter what fixed value on the tangent you take.
对于 “足够好” 的函数，其导数的意义是：无论你用切线原理离 $a$ 多近，结果都是一样的。从这个意义上说，你是对的 —— 你可以在切线上取任何固定值，但 “固定某个值” 不如 “无论在切线上取什么固定值，结果都相同” 更具一般性。

The Core of the Derivative Definition

导数定义的核心

You need to notice that turning $\frac{\Delta y}{\Delta x}$ into $\frac{dy}{dx}$ and approaching one and the same value is in the core of the definition of having a derivative.
你需要注意，将 $\frac{\Delta y}{\Delta x}$ 转化为 $\frac{dy}{dx}$ 并趋近于同一个值，这是导数定义的核心。

Essentially, $\frac{dy}{dx}={\lim }_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$ and that is the way you should understand it.
本质上，$\frac{dy}{dx}={\lim }_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$，这才是你应该理解它的方式。

Conclusion

总结

The derivative “inherits” some (not all) of the behavior of fractions from the fraction (difference quotient) that is at the heart of its definition. That fraction is the motivation for the notation, and the notation makes it easy to remember things like the chain rule, but you have to remember that in spite of all this, it isn’t really a fraction, and you can only treat it as a fraction where there is a theorem that says you can.
我们可以说，导数从其定义核心的分数（差商）中“继承”了一些（并非全部）分数的性质。那个分数是这种符号的设计动机，而且这种符号让人们很容易记住像链式法则这样的内容，但你必须记住，尽管如此，它并不真的是一个分数，只有在有定理表明可以将其当作分数对待的情况下，你才能这样做。

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If $\frac{dy}{dt}dt$ doesn’t cancel, then what do you call it?

如果 $\frac{dy}{dt}dt$ 不能抵消，那么该如何理解它呢？

I have $y$ is a function of $t$. I have reached a situation here where I need to evaluate
我知道 $y$ 是 $t$ 的函数。我遇到了一种情况，需要计算

${\int }_0^b\frac{dy}{dt}dt$

Now clearly $y$ has dependence on $t$, otherwise $\frac{dy}{dt}$ should be 0.
显然，$y$ 依赖于 $t$，否则 $\frac{dy}{dt}$ 应该为 0。

So now I write
所以现在我写成

${\int }_{y(0)}^{y(b)}dy=y{|}_{y(0)}^{y(b)}=y(b)-y(0)$

I know that dt’s don’t cancel_, but what do you call it, then? Just a “change of variables”? How do we justify where $dt$ went?
我知道不能消去，但那么该怎么理解它呢？只是“变量替换”吗？我们如何解释 dt 去哪里了？

edited Apr 13, 2017 at 12:21 CommunityBot
asked Feb 13, 2011 at 18:42 bobobobo

• But now I am also confused at the possibility of the answer not being $y(b)-y(a)$, but from the perspective of differential forms: if $f$ is a differentiable function, then the expression $f^{\prime }(t)dt=(\frac{dy}{dx})dt$ is a differential form (as the differential of a differentiable function), and, in particular, it is an exact form (I am sorry to give this argument, which brings you back into your original question, but in this case, $f^{\prime }(t)dt=d(f(t))$), and, by the fundamental theorem of calc, it should integrate to $f(b)-f(a)$. I don’t see where this argument (basically Stokes’ theorem) could be flawed
  但现在我也对答案可能不是 $y(b)-y(a)$ 感到困惑，但从微分形式的角度来看：如果 $f$ 是可微函数，那么表达式 $f^{\prime }(t)dt=(\frac{dy}{dx})dt$ 是一个微分形式（作为可微函数的微分），特别是，它是一个恰当形式（很抱歉给出这个论点，这会把你带回原来的问题，但在这种情况下，$f^{\prime }(t)dt=d(f(t))$），并且根据微积分基本定理，它的积分应该是 $f(b)-f(a)$。我看不出这个论点（基本上是斯托克斯定理）有什么缺陷

– user7052 Commented Feb 14, 2011 at 9:00

4 Answers

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This is closely related to the recent question_ asking whether $\frac{dy}{dt}$ was a fraction or not, as you note.
正如你所注意到的，这与最近的问题密切相关，该问题询问 $\frac{dy}{dt}$ 是否为分数。

• 微积分 | dy / dx 不是分数吗？-优快云博客
  https://blog.youkuaiyun.com/u013669912/article/details/149782565

As in that question, when Leibnitz first came up with the notation for integrals (the $\int$ sign was really a capital $S$, standing for “summa”, summation), and he wrote ${\int }_a^{b}f(x)dx$, he was really thinking of this as a sum of products, with $dx$ representing an “infinitesimal change in $x$.” He was thinking of dividing the interval $a,b$ into an infinite number of “infinitesimally thin” rectangles, each of height $f(x_i)$ (for whatever $x_i$ you happened to be in), and then adding up all the areas of these infinitesimally thin rectangles. Such a rectangle would have area $f(x_i)dx$ (height times base), and adding them together would yield the area.
与那个问题一样，当莱布尼茨首次提出积分符号时（$\int$ 实际上是大写的 $S$，代表“summa”，即求和），他写下 ${\int }_a^{b}f(x)dx$ 时，实际上是把它看作乘积的和，其中 $dx$ 代表“$x$ 的无穷小变化”。他的想法是将区间 $a,b$ 分成无穷多个“无穷薄”的矩形，每个矩形的高为 $f(x_i)$（无论你恰好处于哪个 $x_i$），然后把这些无穷薄矩形的面积加起来。这样的矩形面积为 $f(x_i)dx$（高乘底），将它们相加就得到总面积。

If you take this point of view (ignoring for a moment the fact that infintesimals can’t exist in the usual real numbers), then the First Fundamental Theorem of Calculus follows by “simple” arithmetic with a telescoping sum (assuming $\frac{dy}{dt}$ is continuous, say). You have that the integral
如果你持这种观点（暂时忽略无穷小量在通常的实数中不存在这一事实），那么微积分第一基本定理可以通过带有 telescoping sum 的“简单”算术得出（假设 $\frac{dy}{dt}$ 是连续的）。你会得到这个积分

${\int }_a^b\frac{dy}{dt}dt$

is really the sum of the quotient of infinitesimal changes in $y$, divided by infinitesimal changes in $t$, multiplied by the infinitesimal change in $t$. If you think of $a,b$ as divided into infinitesimally thin subintervals
实际上是 $y$ 的无穷小变化量与 $t$ 的无穷小变化量的商，再乘以 $t$ 的无穷小变化量的总和。如果你把 $a,b$ 看作被分成无穷薄的子区间

$a=t_0<t_0+dt<t_0+2dt<\cdots <b$

then 那么

$dy=y(t_0+(k+1)dt)-y(t_0+kdt)$,

so that the integral becomes the telescoping sum
因此这个积分就变成了叠缩和（ / 消项和）

$\sum (y(t_0+(k+1)dt)-y(t_0+kdt))=y(b)-y(a)$

because all the “middle terms” cancel out. So here, $dt$ really does cancel out with the $dt$ in $\frac{dy}{dt}$, the sum is telescoping, and the final answer comes out exactly as desired.
因为所有的“中间项”都抵消了。所以在这里，$dt$ 确实与 $\frac{dy}{dt}$ 中的 $dt$ 抵消了，这个和是叠缩和，最终的答案恰好如预期的那样。

But, as with the derivative, there is a legion of logical problems with this way of thinking, not the least of which is that infinitesimals can’t really exist in the usual setting of real numbers. So calculus had to be rewritten. There were proposals by Cauchy on how to define integrals, and eventually we had Riemann’s way of defining integrals as limits. So that when we write ${\int }_a^{b}f(t)dt$ we no longer mean a sum of products of the form $f(t)dt$, but rather we mean a limit of certain Riemann sums. In this view, the “$dt$” is rather more like the symbol to balance the ${\int }_a^b$. Think of the integral sign on the left as a “left parenthesis”, and the $dt$ on the right as the “right parenthesis” that closes out the expression.
但是，与导数一样，这种思维方式存在很多逻辑问题，其中最重要的一点是无穷小量在通常的实数体系中实际上并不存在。因此，微积分必须重新构建。柯西曾提出过积分的定义方法，最终我们有了黎曼将积分定义为极限的方式。所以当我们写下 ${\int }_a^{b}f(t)dt$ 时，我们不再指形如 $f(t)dt$ 的乘积的和，而是指某些黎曼和的极限。在这种观点下，“$dt$”更像是平衡 ${\int }_a^b$ 的符号。可以把左边的积分符号看作“左括号”，右边的 $dt$ 看作结束表达式的“右括号”。

So, just like $\frac{df}{dt}$ no longer literally means “the quotient of an infinitesimal change in $f$ by an infinitesimal change in $t$” but rather means "the limit as $h$ goes to 0 of $(f(t+h)-f(t))/h$, so ${\int }_a^{b}f(t)dt$ no longer literally means “the sum of infinitesimally thin rectangles of height $f(t)$ from $a$ to $b$”, but instead means “the limit of Riemann sums of $f(t)$ over partitions of $a,b$ as the mesh goes to 0”.
所以，就像 $\frac{df}{dt}$ 不再字面上表示“$f$ 的无穷小变化量与 $t$ 的无穷小变化量的商”，而是表示“当 $h$ 趋近于 0 时，$(f(t+h)-f(t))/h$ 的极限”一样，${\int }_a^{b}f(t)dt$ 也不再字面上表示“从 $a$ 到 $b$、高为 $f(t)$ 的无穷薄矩形的和”，而是表示“当网格趋近于 0 时，$f(t)$ 在 $a,b$ 的分割上的黎曼和的极限”。

But one of the great advantages of Leibnitz notation is that it is very suggestive. So you get the First Fundamental Theorem of Calculus, which looks very natural in Leibnitz notation:
但莱布尼茨符号的一大优点是它非常具有启发性。因此你会得到微积分第一基本定理，用莱布尼茨符号表示非常自然：

${\int }_a^b\frac{df}{dt}dt=f(b)-f(a),$

giving the suggestion that you are “cancelling $dt$”, even though you’re not really doing that. But since infinitesimals don’t really exist, this is not literally true. But good notation is not something to be cast aside when it comes along, and Leibnitz notation, being suggestive, is very good notation, so we keep it because it helps with calculations.
这给人一种你在“抵消 $dt$”的暗示，尽管你实际上并没有这样做。但由于无穷小量实际上并不存在，这在字面上并不是真的。但好的符号出现时是不应该被抛弃的，而莱布尼茨符号具有启发性，是非常好的符号，所以我们保留它，因为它有助于计算。

Where did the “$dt$” go? Well, one might ask where the “))” goes in the following calculation:
“$dt$”去哪里了？好吧，有人可能会问在下面的计算中“))”去哪里了：

$2\times (3+5)=16.$

So… where did the “))” go (or where did “$\times$”, “+,” and “((” all go) ? Same place as the “$dt$” went: since it is part of the notation, it “goes away” when we are done with the evaluation.
那么……“))”去哪里了（或者说“$\times$”、“+”和“(("都去哪里了）？和“$dt$”去的地方一样：因为它是符号的一部分，当我们完成计算时，它就“消失”了。

Note. As with derivatives, with Nonstandard Analysis one can write calculus so that the $dt$ in the integral really represents a quantity you are multiplying by and then adding, so that in nonstandard analysis the First Fundamental Theorem of Calculus really is just the observation that if you divide by $dx$ and multiply by $dx$, then the two cancel out.
注： 与导数一样，在非标准分析中，我们可以这样表述微积分：积分中的 $dt$ 实际上代表一个你要相乘然后相加的量，因此在非标准分析中，微积分第一基本定理实际上只是这样一个观察结果：如果你除以 $dx$ 再乘以 $dx$，那么这两个 $dx$ 就抵消了。

edited Apr 13, 2017 at 12:21 Community Bot
answered Feb 13, 2011 at 19:46 Arturo Magidin

• @bobobobo: I don’t personally like Wolfram alpha, and I certainly would not link to it. So kindly do not make it appear that I did.
  @bobobobo：我个人不喜欢 Wolfram Alpha，当然也不会链接到它。所以请不要让人觉得我链接了。

– Arturo Magidin Commented Mar 18, 2011 at 16:19

bobobobo: I think if you use the accurate expression $dy=y^{\prime }(x)dx$–where $y^{\prime }(x)$ is the limit quotient–at least some of the ambiguity will go away.
bobobobo：我认为如果你使用准确的表达式 $dy=y^{\prime }(x)dx$——其中 $y^{\prime }(x)$ 是极限商——至少一些歧义会消失。

– gary Commented Apr 25, 2011 at 23:26

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This is where the Fundamental theorem of calculus comes in handy.
这就是微积分基本定理派上用场的地方。

It comes in two flavors:
它有两种形式：

The first flavor (Riemann Setting)
第一种形式（黎曼体系）

Let $f(x)$ be a continuous function in $x$ on an interval $a,b$.
设 $f(x)$ 是区间 $a,b$ 上关于 $x$ 的连续函数。

$F(x)={\int }_{a_0}^{x}f(y)dy$, where $a_0\in [a,b]$ is called a primitive of $f(x)$.
$F(x)={\int }_{a_0}^{x}f(y)dy$，其中 $a_0\in [a,b]$，称为 $f(x)$ 的原函数。

(The existence of a primitive can be proved since $f(x)$ is continuous.)
（由于 $f(x)$ 是连续的，原函数的存在性可以得到证明。）

Then the Fundamental theorem of calculus states that
那么微积分基本定理表明

$F(x)\in C^1(a,b)\frac{dF(x)}{dx}=f(x)$

The second flavor (Riemann Setting)
第二种形式（黎曼体系）

Let $f(x)$ be a continuous function on an interval $a,b$ and let its derivative $f^{\prime }(x)$ exist on $(a,b)$ and let $f^{\prime }(x)$ be integrable.
设 $f(x)$ 是区间 $a,b$ 上的连续函数，其导数 $f^{\prime }(x)$ 在 $(a,b)$ 上存在且 $f^{\prime }(x)$ 可积。

Then the Fundamental theorem of calculus states that
那么微积分基本定理表明

${\int }_{a_0}^{b_0}\frac{df(y)}{dy}dy=f(b_0)-f(a_0)$

where
其中

$a_0,b_0\in [a,b]$

The second flavor is the one you use in your argument above.
第二种形式就是你在上面的论证中所使用的。

You might be able to relax some conditions that I have stated to obtain the same conclusion.
你或许可以放宽我所陈述的一些条件，以得到相同的结论。

Again, this is a place where a good notation helps. $\frac{dy}{dt}dt$ can be “treated” like usual ratio and the $dt$’s can be “canceled” out though the actual reasoning for the “canceling” out is different.
再次强调，这是一个好的符号能起到帮助作用的地方。$\frac{dy}{dt}dt$ 可以像通常的比例一样被“处理”，并且 $dt$ 可以被“抵消”，尽管“抵消”的实际推理是不同的。

EDIT

As I always believe counterexamples are a great way to study a theorem or a property. Below are examples where you cannot “cancel out the $dt$’s”.
正如我一直认为的，反例是研究定理或性质的好方法。下面是一些你不能“抵消 $dt$”的例子。

In case of Riemann Integration, if you take $y(t)$ as the Volterra function (Thanks to Theo Buehler for pointing that one), then $y(t)$ is differentiable i.e. $\frac{dy}{dt}$ exists for all $t$. However, $\frac{dy}{dt}$ is not Riemann integrable i.e.
在黎曼积分的情况下，如果你取 $y(t)$ 为沃尔泰拉函数（感谢 Theo Buehler 指出这一点），那么 $y(t)$ 是可微的，即 $\frac{dy}{dt}$ 对所有 $t$ 都存在。然而，$\frac{dy}{dt}$ 不 是黎曼可积的，即

${\int }_a^b{y}^{\prime }(t)dt$

doesn’t exists where the integral is interpreted in Riemann sense.
在积分被理解为黎曼积分的情况下是不存在的。

It is however, Lebesgue integrable.
然而，它是勒贝格可积的。

In case of Lebesgue Integration, there exists continuous functions, $f(x)$, which are differentiable almost everywhere i.e. $f^{\prime }(x)$ exists almost everywhere, and its derivative, $f^{\prime }(x)$, is Lebesgue integrable but the Lebesgue integral of $f^{\prime }(x)$ is not equal to the change in $f(x)$ i.e.
在勒贝格积分的情况下，存在连续函数 $f(x)$，它们几乎处处可微，即 $f^{\prime }(x)$ 几乎处处存在，且其导数 $f^{\prime }(x)$ 是勒贝格可积的，但 $f^{\prime }(x)$ 的勒贝格积分不等于 $f(x)$ 的变化量，即

${\int }_a^b{f}^{\prime }(x)dx\ne f(b)-f(a)$

where the integral is interpreted in Lebesgue sense.
其中积分被理解为勒贝格积分。

The famous function here is the Cantor function $C(x)$. It is continuous everywhere and has zero derivative almost everywhere. Hence, the Lebesgue integral ${\int }_0^1{C}^{\prime }(x)dx$ in the interval $[0,1]$ is $0$. However, $C(1)-C(0)=1$
这里著名的函数是康托尔函数 $C(x)$。它处处连续，且几乎处处导数为零。因此，在区间 $[0,1]$ 上，勒贝格积分 ${\int }_0^1{C}^{\prime }(x)dx$ 为 $0$。然而，$C(1)-C(0)=1$

edited Feb 14, 2011 at 17:08
answered Feb 13, 2011 at 18:54

user17762

• @Sivaram: Please check the beginning of the second part.
  @Sivaram：请检查第二部分的开头。

– Shai Covo Commented Feb 13, 2011 at 21:47

• @Shai Covo: Could you explain your comment? Feel free to edit if there are some minor changes to be done.
  @Shai Covo：你能解释一下你的评论吗？如果有一些小的修改需要做，请随意编辑。

– user17762 Commented Feb 13, 2011 at 21:58

• @Shai Covo: True the wiki link says so. But I am wondering if it follows that $f^{\prime }$ is integrable, whenever $f^{\prime }$ is the derivative of a continuous $f$. Could you explicitly give me an example which counters this?
  @Shai Covo：维基链接确实是这么说的。但我想知道，当 $f^{\prime }$ 是连续函数 $f$ 的导数时，是否可以推出 $f^{\prime }$ 是可积的。你能明确给我一个反例吗？

– user17762 Commented Feb 13, 2011 at 22:09

If you’re asking about Riemann integrability, one standard example is Volterra’s function. You might want to check out these nice slides: macalester.edu/~bressoud/talks/AlleghenyCollege/Wrestling.pdf On the other hand, in the context of Lebesgue integration you have Lebesgue’s differentiation theorem characterizing the absolutely continuous functions as precisely those for which the fundamental theorem of calculus holds.
如果你问的是黎曼可积性，一个标准的例子是沃尔泰拉函数。你可能想看看这些不错的幻灯片：macalester.edu/~bressoud/talks/AlleghenyCollege/Wrestling.pdf（链接已沉寂） 另一方面，在勒贝格积分的背景下，你会接触到勒贝格微分定理，该定理将 绝对连续 函数刻画为恰好是那些使微积分基本定理成立的函数。

– t.b. Commented Feb 13, 2011 at 23:47

There is also $f(x)=x^2\sin (\frac{1}{x^2})$, with $f(0)=0$; this $f$ is differentiable everywhere, but its derivative is not even Lebesgue integrable on any interval containing $0$. (This is possible because the derivative is unbounded.) Yet even so, it is still Henstock-integrable. Every derivative is Henstock-integrable, and the integral recovers the original function up to a constant (if the domain is an interval).
还有 $f(x)=x^2\sin (\frac{1}{x^2})$，其中 $f(0)=0$；这个 $f$ 处处可微，但其导数在任何包含 $0$ 的区间上甚至都不是勒贝格可积的。（这是可能的，因为导数是无界的。）然而，即便如此，它仍然是亨斯托克可积的。每个 导数都是亨斯托克可积的，并且积分可以恢复原函数（相差一个常数，如果定义域是一个区间的话）。

– Toby Bartels Commented Apr 29, 2017 at 6:54

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via:

• Is this a correct/good way to think interpret differentials for the beginning calculus student? - Mathematics Stack Exchange - 2015
  https://math.stackexchange.com/questions/1548487/is-this-a-correct-good-way-to-think-interpret-differentials-for-the-beginning-ca/1590019#1590019

• calculus - If $\frac{dy}{dt}dt$ doesn’t cancel, then what do you call it? - Mathematics Stack Exchange - 2011
  https://math.stackexchange.com/questions/21869/if-fracdydtdt-doesnt-cancel-then-what-do-you-call-it

• 微积分 | dy / dx 不是比率吗？-优快云博客
  https://blog.youkuaiyun.com/u013669912/article/details/149719364