Investigation 3 – The Product Rule | 探究3：乘积法则

📚 Investigation 3 – The Product Rule | 探究3：乘积法则

In previous investigations we learned how to differentiate simple functions and sums of functions. But what happens when two differentiable functions are multiplied together? Does the derivative of a product simply equal the product of the derivatives? In this investigation you will gather data, formulate a conjecture, and ultimately derive one of the most important rules of calculus: the product rule.

在之前的探究中，我们学习了如何对简单函数以及函数的和进行求导。但是，当两个可微函数相乘时会发生什么呢？乘积的导数是否就等于各导数的乘积？在本次探究中，你将收集数据、提出猜想，并最终推导出微积分中最重要的法则之一：乘积法则。

1. Introduction: The Problem of Differentiating Products | 引言：乘积求导的问题

we deal with sums or differences of functions, differentiation is straightforward: (f + g)' = f' + g'. However, many real-world and mathematical models involve products, such as h(x) = x² sin x or h(x) = e^x ln x. If we naively multiply the individual derivatives, we get 2x · cos x and e^x · (1/x), neither of which is correct. This investigation aims to uncover the actual pattern behind differentiating products.

当我们处理函数的和或差时，求导非常直接：(f + g)' = f' + g'。然而，许多现实世界和数学模型都涉及乘积，例如 h(x) = x² sin x 或 h(x) = e^x ln x。如果我们天真地将各导数相乘，会得到 2x · cos x 和 e^x · (1/x)，它们都是不正确的。本探究旨在揭示乘积求导背后的真正模式。

2. Revisiting the Definition of the Derivative | 重温导数定义

All differentiation rules stem from the limit definition of the derivative. For a function f(x), the derivative at a point x is defined as
f'(x) = lim_h→0 (f(x+h) - f(x))/h, provided the limit exists.
This first-principles approach will be essential for proving any new rule we discover. In this investigation we will use Δx as the small change rather than h, writing
f'(x) = lim_Δx→0 (f(x+Δx) - f(x))/Δx.

所有求导法则都源于导数的极限定义。对于函数 f(x)，在点 x 处的导数定义为
f'(x) = lim_h→0 (f(x+h) - f(x))/h，只要该极限存在。
这一第一原理方法将对我们证明所发现的任何新法则至关重要。在本次探究中，我们将使用 Δx 作为微小变化量（而不是 h），写作
f'(x) = lim_Δx→0 (f(x+Δx) - f(x))/Δx。

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3. Exploring Simple Polynomial Products | 探索简单的多项式乘积

Let us investigate several examples where we already know the derivative by expanding the product first. Complete the table below to compare the true derivative with the product of individual derivatives. The final column tests a combination that appears in many textbooks: the derivative of the first times the second plus the first times the derivative of the second.

让我们考察几个例子，我们先展开乘积再进行求导，从而已知其导数。完成下表，将真正的导数与各导数之积进行比较。最后一列测试了一个在许多教材中出现的组合：第一项的导数乘第二项加上第一项乘第二项的导数。

From the table, the column f'(x)g'(x) clearly fails to match the true derivative. However, the combination f'(x)g(x) + f(x)g'(x) perfectly reproduces h'(x) in every case. This suggests a remarkable pattern.

从表格中可以清楚看出，f'(x)g'(x) 这一列与真实导数不符。然而，组合 f'(x)g(x) + f(x)g'(x) 在每一种情况下都完美地重现了 h'(x)。这揭示了一个极好的模式。

4. Spotting the Pattern: A Conjecture | 发现模式：猜想

hen the derivative of their product is not the product of the derivatives, but rather (uv)' = u'v + uv' same and multiply by the derivative of the first, and finally add the two terms together.

基于这些数值证据，我们可以提出猜想：如果 u 和 v 是可微函数，那么它们乘积的导数并不是导数的乘积，而是 (uv)' = u'v + uv'。用文字表达就是：保持第一个函数不变，乘上第二个函数的导数；再保持第二个函数不变，乘上第一个函数的导数；然后将这两项相加。

5. The Product Rule – Formal Statement | 乘积法则的正式表述

en the product function h(x) = u(x)v(x) is differentiable, and its derivative is given by:

h'(x) = u'(x)v(x) + u(x)v'(x)

In Leibniz notation, if y = uv, where u and v are functions of x, then:

dy/dx = (du/dx) v + u (dv/dx)

设 u(x) 和 v(x) 为可微函数。那么乘积函数 h(x) = u(x)v(x) 可微，且其导数为：

h'(x) = u'(x)v(x) + u(x)v'(x)

使用莱布尼茨记号，如果 y = uv，其中 u 和 v 都是 x 的函数，则：