微积分 | 导数概念 / 定义 / 符号（一些久远的讨论）

注：本文为 “微积分 | 导数” 相关讨论合辑。
英文引文，机翻未校。
如有内容异常，请看原文。

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csdn 篇幅所限， 另一篇

• 微积分 | 导数概念 / 定义 / 符号（一些讨论）-优快云博客
  https://blog.youkuaiyun.com/u013669912/article/details/149739083

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Differentials

微分

Date: 09/18/98 at 01:09:03
From: Maria sanabria
Subject: Differentials
主题：微分

I have to reach this conclusion:

我需要得出这样的结论：

If you can get the differentials of a function, you can differentiate it, but if you can differentiate it, you can not necessarily get its differentials.
Please help.

若能得到一个函数的微分，则必能对其求导；但能对一个函数求导，却未必能得到其微分。
请帮忙。

Date: 09/18/98 at 07:35:27
From: Doctor Jerry
Subject: Re: Differentials

Hi Maria,

嗨，Maria，

The standard definition of the differential of a real - valued function $f$ of a real variable is:

实值函数 $f$（以实变量为自变量）的微分的标准定义如下：

At a given point $x$, the differential $d{f}_x$ (df sub x; usually the $x$ is omitted) of $f$ is the linear function defined on $\mathbb{R}$ by:

在给定点 $x$ 处，$f$ 的微分 $d{f}_x$（通常省略下标 $x$）是在 $\mathbb{R}$ 上定义的线性函数，定义为：

$d{f}_x(h)=f^{\prime }(x)\times h$

Everyday usage of the differential often suppresses the fact that the differential is a linear function.

日常使用中，微分通常被当作一个线性函数，但这一点往往被忽视。

For example, if $y=f(x)=x^2$, then we write:

例如，如果 $y=f(x)=x^2$，那么我们写作：

$dy=df=2x\times dx$

where $dx$ is used instead of $h$.

其中 $dx$ 用来代替 $h$。

This is for good reason.

这是有原因的。

The finite numbers $dy$ and $dx$ appearing in $dy=2x\times dx$ can be manipulated to obtain:

在 $dy=2x\times dx$ 中出现的有限数 $dy$ 和 $dx$ 可以被操作以得到：

$\frac{dy}{dx}=2x$

I feel that I haven’t replied directly to your question.

我觉得我没有直接回答你的问题。

I think that this is because I don’t fully understand your question.

我想这是因为我没有完全理解你的问题。

Please write again if my answer has not helped.

如果我的回答没有帮助到你，请再次写信。

Date: 09/19/98 at 13:57:17
From: Maria Stella Sanabria
Subject: Differentials

Thanks for your answer.

谢谢你的回答。

I know that the question is a little bit confusing, and at the beginning I thought it was a problem of the translation from English of the Math books.

我知道这个问题有点令人困惑，一开始我以为这是数学书从英文翻译过来的问题。

Your answer helped a little, so I am going to try to rephrase it.

你的回答有点帮助，所以我要尝试重新表述一下。

What is the difference between finding the derivatives of a function (dy/dx), and finding its differentials (dy, dx)?

求函数的导数（dy/dx）和求它的微分（dy, dx）有什么区别？

In the books I’ve seen they define differentials supposing that $f(x)$ is differentiable.

在我看过的书中，他们都是假设 $f(x)$ 可微的情况下定义微分的。

My teacher gave a hint to reach this conclusion: if you can find the differentials of $f$, then $f$ is differentiable, but if $f$ is differentiable you can’t necessarily find its differentials.

我的老师给出了这样一个提示：如果你能找到 $f$ 的微分，那么 $f$ 就是可微的，但是如果 $f$ 是可微的，你却不一定能找到它的微分。

That is why I can prove this, starting with a function that is differentiable.

这就是为什么我可以从一个可微的函数开始来证明这一点。

Thanks.

谢谢。

Date: 09/22/98 at 08:53:36
From: Doctor Jerry
Subject: Re: Differentials

Hi Maria,

嗨，Maria，

Suppose $f(x)=x^2$.

假设 $f(x)=x^2$。

To find the derivative of $f$ we use the definition of derivative: $f^{\prime }(x)$ is the limit as $h\to 0$ of the quotient

为了求 $f$ 的导数，我们使用导数的定义：$f^{\prime }(x)$ 是当 $h\to 0$ 时，商的极限

$\frac{f(x+h)-f(x)}{h}$

For this function, $f^{\prime }(x)=2x$.

对于这个函数，$f^{\prime }(x)=2x$。

Okay, this much is clear; there is no possible ambiguity.

好的，这一点是明确的，没有任何可能的歧义。

The differential of $f$ at $x$ is defined to be the linear function $df$, which is defined on all of $\mathbb{R}$ by:

在 $x$ 处，$f$ 的微分被定义为线性函数 $df$，它在 $\mathbb{R}$ 上定义为：

$df(h)=f^{\prime }(x)\times h$

Often, the notation $df(h)$ is shortened to $df$ or, if $y=f(x)$, then we write $dy$ instead of $df$.

通常，$df(h)$ 的记法被简化为 $df$，或者如果 $y=f(x)$，那么我们用 $dy$ 来代替 $df$。

Then the above definition is:

那么上述定义就是：

$dy=f^{\prime }(x)\times dx$ or

或者

$\frac{dy}{dx}=f^{\prime }(x)$

Unless you are studying differential geometry, in which $dx$ is interpreted slightly differently, $dx$ is not the differential of a function.

除非你正在学习微分几何，在那里 $dx$ 的解释略有不同，否则 $dx$ 不是函数的微分。

It is a variable, the same as $h$.

它是一个变量，和 $h$ 一样。

I think the definitions you found in the books and what I said above are comparable.

我认为你在书中找到的定义和我上面说的是一致的。

However, in elementary calculus the definitions change a bit when one goes from functions of one variable to functions of two or more variables.

然而，在初等微积分中，当从单变量函数过渡到多变量函数时，定义会有一些变化。

In one variable it doesn’t matter whether one defines the derivative of a function first, as I did above, and then defines the differential (the linear function idea) or does the reverse (see below).

在单变量的情况下，先定义函数的导数（如我上面所做），然后定义微分（线性函数的概念），或者反过来（见下文），这并不重要。

They are equivalent.

它们是等价的。

For functions of more than one variable, however, big differences become evident.

然而，对于多变量函数，大的差异变得明显。

Here is the definition of differentiability that works for functions of one or more variables.

这里是一个适用于一个或多个变量函数的可微性定义。

Suppose that $f$ is a function on $(a,b)$ and $p$ is a point of $(a,b)$.

假设 $f$ 是 $(a,b)$ 上的一个函数，$p$ 是 $(a,b)$ 中的一个点。

Then $f$ is differentiable at $p$ if a real number $P$ can be found for which

那么如果能找到一个实数 $P$，使得

$\frac{|f(p+h)-f(p)-h\times P|}{|h|}\to 0$ as $h\to 0$

$\frac{|f(p+h)-f(p)-h\times P|}{|h|}\to 0$ 当 $h\to 0$

Suppose that $f$ is differentiable at $p$, $P$ is the derivative of $f$ at $p$ and we write $f^{\prime }(p)=P$.

假设 $f$ 在 $p$ 处可微，$P$ 是 $f$ 在 $p$ 处的导数，我们写作 $f^{\prime }(p)=P$。

Also, the function $L(h)=h\times P$, from $\mathbb{R}$ into $\mathbb{R}$, is called the differential of $f$ at $p$, often written as $df$.

另外，函数 $L(h)=h\times P$，从 $\mathbb{R}$ 映射到 $\mathbb{R}$，被称为 $f$ 在 $p$ 处的微分，通常写作 $df$。

The differential is a mapping; the derivative is a number.

微分是一个映射；导数是一个数。

This is equivalent to the earlier definition of derivative, that is, if one assumes either, the other follows.

这等同于前面的导数定义，也就是说，如果假设了其中一个，另一个就会随之而来。

Now, this form carries over to functions of two (and more) variables.

现在，这种形式延续到了两个（及更多）变量的函数。

Suppose that $f$ is a function defined on the set $I$ of all $(x,y)$ for which $a<x<y$ and $c<y<d$ and $(p,q)$ is a point of $I$.

假设 $f$ 是在集合 $I$ 上定义的函数，对于所有 $(x,y)$，$a<x<y$ 且 $c<y<d$，$(p,q)$ 是 $I$ 中的一个点。

Then $f$ is differentiable at $(p,q)$ if real numbers $P$ and $Q$ can be found for which:

那么如果能找到实数 $P$ 和 $Q$，使得：

$\frac{|f(p+h,q+k)-f(p,q)-h\times P-k\times Q|}{\sqrt{h^2+k^2}}\to 0$ as $(h,k)\to (0,0)$

$\frac{|f(p+h,q+k)-f(p,q)-h\times P-k\times Q|}{\sqrt{h^2+k^2}}\to 0$ 当 $(h,k)\to (0,0)$

If $f$ is differentiable, then $f$ has partial derivatives at $(p,q)$, but the converse is not true.

如果 $f$ 是可微的，那么 $f$ 在 $(p,q)$ 处有偏导数，但反之则不然。

• Doctor Jerry, The Math Forum

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Can dy/dx Be Treated as a Fraction?

 
$\frac{dy}{dx}$ 能否被当作分数处理？

Date: 08/26/2004 at 14:57:25
From: Amit
Subject: Properties of $\frac{dy}{dx}$

主题：$\frac{dy}{dx}$ 的性质

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，我了解到 $\frac{dy}{dx}$ 是一个表示“$y$ 关于 $x$ 的导数”的符号。我理解了这一点。
但我对 $\frac{dy}{dx}$ 是否可以被当作分数处理感到困惑，即是否可以分别赋予 $dy$ 和 $dx$ 各自的意义。

For example, in integration by substitution:

例如，在积分替换中：

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

$u=3x+5$

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

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

That is the part that confuses me. How can the $dx$ be solved for? What exactly is " $dx$ "?

这就是让我感到困惑的部分。$dx$ 怎么能被解出来呢？“$dx$”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…

我明白 $\frac{dy}{dx}$ 是一个极限，它是一个斜率。但仅仅把它当作一个符号并不能帮助我理解如何将分母展开。希望得到一些帮助……

Date: 08/26/2004 at 15:27:40
From: Doctor Vogler
Subject: Re: Properties of $\frac{dy}{dx}$

Hi Amit,

嗨，Amit，

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,

感谢你给“Dr. Math”写信。对你问题的简单回答是，你对 $\frac{dy}{dx}$ 的定义是正确的；它表示 $y$ 关于 $x$ 的导数，而单独写的 $dy$ 和 $dx$ 是没有意义的，

so that

因此

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

is not a meaningful expression but should be written

不是一个有意义的表达式，而应该写作

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

And when certain nice things happen that look like fractions, such as:

当出现一些看起来像分数的“好”情况时，例如：

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

and

以及

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

then this is actually just the Chain Rule at work.

这实际上只是链式法则在起作用。

And the reason that

而其原因

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

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.

而是链式法则再次起作用。

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.

所有这些都是正确的，除了我应该对“不是一个有意义的表达式”进行限定。你看，某样东西只是在有人赋予它正式的意义之前是没有意义的。然后你希望他们赋予它的意义具有有用的性质（例如，它与导数有关……）。事实上，这已经完成了，并且有许多数学知识被用于微分理论，它适用于积分，并且根据这一理论，在每个积分的末尾加上微分“$dx$”也是有意义的，等等。

One math doctor alluded to some of this on

一位数学博士在以下链接中提到了一些相关内容：

• Differentials - Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/

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.

你也可以找到更详细讨论这一内容的书籍。但事实是，大多数使用微积分的人并不真正需要所有的微分理论，而链式法则确实足以验证你从将 $\frac{dy}{dx}$ 作为分数处理中得到的大多数事实。

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.

你可以将它当作分数处理的原因是 $\frac{dy}{dx}$ 是一个分数的极限，因此你可以在取极限之前对这个分数进行大多数操作。换句话说，在取极限之前，它是一个分数，所以你可以把它当作一个分数来处理。但随后你取了极限，它就变成了一个导数。

Finally, there is also the theory of estimating with derivatives, where I always say to think of

最后，还有使用导数进行估算的理论，在这里我总是说要这样想：

$dx=$ change in $x$
$dx=$ $x$ 的变化量

$dy=$ change in $y$
$dy=$ $y$ 的变化量

$x=$ unchanged value of $x$
$x=$ $x$ 的不变值

$y=$ unchanged value of $y$
$y=$ $y$ 的不变值

For example, to estimate $(1.98{)}^6$, we use
例如，为了估算 $(1.98{)}^6$，我们使用

$y=x^6$
$dy=6x^{5}dx$
$x=2$
$dx=-0.02$

(so that $x+dx=1.98$), and therefore
（使得 $x+dx=1.98$），因此
$y=2^6=64$
$dy=6x^{5}dx=6\times 32\times (-0.02)=-3.84$

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 $(1.98{)}^6$ 非常接近。这本质上是在取极限之前将 $\frac{dy}{dx}$ 作为分数处理，因为 $dx$ 并没有完全趋向于零，而是趋向于 $-0.02$。

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.

这是否有助于你理解微分在某种意义上是分数，但在其他意义上又不是分数？如果你对此有任何疑问或需要更多帮助，请回信并告诉我你已经做了什么，我会尽力提供进一步的建议。

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Why Does Integration by Substitution Work?

为什么积分替换法有效？

Date: 12/13/2006 at 22:03:56
From: Reuben
Subject: Why integration by substitution works

主题：为什么积分替换法有效

I just learned the process of integration by substitution. My textbook seems to explain it for the most part, but it says “imagine that you multiply $\frac{du}{dx}$ by the $dx$ in the derivative to get $du$.” This seems to work, but I’ve found online that technically, you can’t treat differentials as numbers that can be “canceled out.”

我刚刚学习了积分替换法的过程。我的教科书似乎在很大程度上解释了它，但它说“想象一下，你将 $\frac{du}{dx}$ 乘以导数中的 $dx$ 以得到 $du$”。这似乎行得通，但我在网上发现，从技术上讲，你不能把微分当作可以“抵消”的数字。

A) Is this true? It seems to me that they’re just very very very small numbers. So why could you not cancel them out?

A) 这是真的吗？在我看来，它们只是非常非常非常小的数字。那么为什么你不能抵消它们呢？

B) If that’s the case, how do you prove that substitution works and use integration by substitution WITHOUT treating differentials as numbers?

B) 如果是这样的话，你如何证明替换法有效，并且在不把微分当作数字的情况下使用积分替换法呢？

Date: 12/14/2006 at 08:39:02
From: Doctor Fenton
Subject: Re: Why integration by substitution works

主题：回复：为什么积分替换法有效

Hi Reuben,
嗨，Reuben，

Thanks for writing to Dr. Math. There are several ways to look at differentials, but probably the simplest is to view them as a formal bookkeeping device for keeping track of the constants when finding antiderivatives.

感谢你给“Dr. Math”写信。微分有几种不同的看待方式，但最简单的可能是把它们当作一种形式化的簿记工具，用于在寻找原函数时跟踪常数。

Many of the integration (or antidifferentiation) rules are actually counterparts of corresponding differentiation rules, and this is true of the substitution theorem, which is the integral version of the Chain Rule.

许多积分（或求原函数）规则实际上是相应微分规则的对应物，而替换定理就是链式法则的积分版本。

The Chain Rule says that if we have a composite function $F(g(x))$, and if $f(x)=F^{\prime }(x)$ is the derivative of the outer function $F(x)$, then

链式法则指出，如果我们有一个复合函数 $F(g(x))$，并且如果 $f(x)=F^{\prime }(x)$ 是外函数 $F(x)$ 的导数，那么

$[F(g(x)){]}^{\prime }=F^{\prime }(g(x))\times g^{\prime }(x)$
$=f(g(x))\times g^{\prime }(x)$.

The antiderivative version of this says that if we want to find the antiderivative of $f(g(x))$ and we know the antiderivative $F$ of $f$, then the antiderivative of $f(g(x))$ is just $F(g(x))$, and we have reduced the problem of finding the antiderivative of the complicated expression $f(g(x))$ to that of finding the antiderivative of $f$, which we usually write with a different independent variable such as $u$. That is, the Chain Rule says that

这个法则的原函数版本指出，如果我们想要求 $f(g(x))$ 的原函数，并且我们知道 $f$ 的原函数 $F$，那么 $f(g(x))$ 的原函数就是 $F(g(x))$，我们已经将求复杂表达式 $f(g(x))$ 的原函数的问题简化为求 $f$ 的原函数，我们通常用另一个独立变量，比如 $u$，来表示它。也就是说，链式法则指出

$[F(g(x)){]}^{\prime }=f(g(x))\times g^{\prime }(x)$ (remember that $F^{\prime }=f$)

$[F(g(x)){]}^{\prime }=f(g(x))\times g^{\prime }(x)$（记住 $F^{\prime }=f$）

so

因此

$\int f(g(x))\times g^{\prime }(x)dx=F(g(x))+C$.

But since $F^{\prime }=f$,

但由于 $F^{\prime }=f$，

$\int f(u)du=F(u)$

(the letter used for the variable of integration is a dummy variable, so we can use any letter we wish). This lets us write

（积分变量的字母是一个虚拟变量，因此我们可以使用任何字母）。这让我们可以写作

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

where we understand the right side to be $F(u)$ with $u$ replaced by the formula $g(x)$. So it appears that in the integral on the left side, we replaced $g(x)$ by $u$ and $g^{\prime }(x)dx$ by $du$ (similar to the way it “appears” that the sun rises and sets). This process doesn’t have to be considered meaningful in itself, but rather just a mnemonic or aid in determining the outer function $f(x)$ in the composition. The differential part of this substitution can also be thought of as a mnemonic, since we can mix Leibniz and function notation to write, if $u=g(x)$, that

我们理解右边是 $F(u)$，其中 $u$ 被替换为公式 $g(x)$。因此，在左边的积分中，我们似乎将 $g(x)$ 替换为 $u$，将 $g^{\prime }(x)dx$ 替换为 $du$（类似于“出现”太阳升起和落下的方式）。这个过程本身不一定需要被视为有意义的，而只是作为助记符或帮助确定复合函数中外函数 $f(x)$ 的辅助工具。这种替换的微分部分也可以被视为一种助记符，因为我们可以混合使用莱布尼茨和函数符号来写作，如果 $u=g(x)$，那么

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

and “multiplying” by $dx$ gives $du=g^{\prime }(x)dx$. One can make this procedure logically rigorous by introducing the concept of differential forms, but that requires a lot of mathematical machinery, so that it is easier to just think of it as a formal computation. (There are books such as Bressoud’s Second Year Calculus, Edwards’ Advanced Calculus: A Differential Forms Approach, and Spivak’s Calculus on Manifolds which go over this in detail.)

并且“乘以” $dx$ 得到 $du=g^{\prime }(x)dx$。可以通过引入微分形式的概念来使这个过程逻辑上严格，但这需要大量的数学工具，因此更容易地只是将其视为一种形式化的计算。（有一些书籍，如 Bressoud 的《第二年微积分》、Edwards 的《高级微积分：微分形式方法》和 Spivak 的《流形上的微积分》，详细讨论了这一点。）

For example, to integrate

例如，为了积分

$\int [x^3+1{]}^{1/2}\times x^{2}dx$,

we can notice that the integrand has the form of the derivative of a composite function $f(g(x))\times g^{\prime }(x)$, with $f(x)=x^{1/2}$ and $g(x)=x^3+1$, but $g^{\prime }(x)=3x^2$, not just $x^2$, so the constant is not quite right. However, if we multiply and divide by 3, we have

我们可以注意到，被积函数具有复合函数 $f(g(x))\times g^{\prime }(x)$ 的导数形式，其中 $f(x)=x^{1/2}$，$g(x)=x^3+1$，但 $g^{\prime }(x)=3x^2$，而不是仅仅是 $x^2$，因此常数项不太正确。然而，如果我们乘以并除以 3，我们得到

$\int [x^3+1{]}^{1/2}\times x^{2}dx=\frac{1}{3}\int [x^3+1{]}^{1/2}\times (3x^2)dx$

and the integrand on the right is exactly a derivative now, of the composite function $\frac{2}{3}[x^3+1{]}^{3/2}$ (we can see this because the outer function $f(x)=x^{1/2}$， whose antiderivative $F(x)=\frac{2}{3}{x}^{3/2}$， so the antiderivative is $\frac{1}{3}F(g(x))$）。This argument has computed the integral or antiderivative without substitution.

右边的被积函数现在正好是一个复合函数 $\frac{2}{3}[x^3+1{]}^{3/2}$ 的导数（我们可以看到这一点，因为外函数 $f(x)=x^{1/2}$ 的原函数 $F(x)=\frac{2}{3}{x}^{3/2}$，因此原函数是 $\frac{1}{3}F(g(x))$）。这个论证在没有使用替换的情况下计算了积分或原函数。

To use substitution, we let $u=x^3+1$, and $du=3x^{2}dx$, so that $x^{2}dx=\frac{1}{3}du$, and

为了使用替换，我们令 $u=x^3+1$，并且 $du=3x^{2}dx$，因此 $x^{2}dx=\frac{1}{3}du$，并且

$\int [x^3+1{]}^{1/2}\times x^{2}dx=\int u^{1/2}\times \frac{1}{3}du$

$=\frac{1}{3}\times \frac{2}{3}{u}^{3/2}+C$

$=\frac{2}{9}[x^3+1{]}^{3/2}+C$.

Substitution makes the process fairly mechanical so it doesn’t require much thought, once you see the appropriate substitution to use, and it also automatically keeps the constants straight.

替换法使这个过程相当机械化，因此不需要太多思考，一旦你看到了合适的替换方法，并且它也能自动地保持常数的正确性。

The objective of indefinite integration is to find an antiderivative, and exactly how you do that isn’t really important, at least in my opinion. If you can just look at a formula and “see” what the antiderivative is, you have solved the problem. Most of us can’t do that, and there are a number of procedures to help, such as the substitution rule, and integration by parts (the integral version of the Product Rule of differentiation). The real theory behind substitution is the Chain Rule, and you can look at the details of substitution as a formal process for helping you see the important parts of the composite functions involved, without worrying about their intrinsic meaning.

不定积分的目标是找到一个原函数，而你如何做到这一点并不真的重要，至少在我看来是这样。如果你能仅仅看一个公式就能“看出”原函数是什么，那么你就解决了问题。我们大多数人做不到这一点，而且有一些程序可以帮助，比如替换规则和分部积分法（微分的乘积法则的积分版本）。替换背后的真正理论是链式法则，你可以把替换的细节看作是一个形式化的程序，帮助你看到所涉及的复合函数的重要部分，而不必担心它们的内在含义。

If you have any questions, please write back and I will try to explain further.

如果你有任何问题，请回信，我会尽力进一步解释。

• Doctor Fenton, The Math Forum

Date: 12/14/2006 at 13:45:51
From: Reuben
Subject: Thank you (Why integration by substitution works)

主题：感谢（为什么积分替换法有效）

Thank you so much. Your response has made the reasoning behind this SO much clearer to me!

非常感谢！你的回答让我对这个背后的推理清晰多了！

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via：

• Differentials - Math Forum - Ask Dr. Math
  http://mathforum.org/library/drmath/view/53678.html

• Can dy/dx Be Treated as a Fraction? - Math Forum - Ask Dr. Math
  http://mathforum.org/library/drmath/view/65462.html

• Why Does Integration by Substitution Work? - Math Forum - Ask Dr. Math
  http://mathforum.org/library/drmath/view/69805.html