📚 The Product Rule: Differentiating Products of Two Functions | 乘积法则：两个函数相乘的求导
When two functions are multiplied together, their derivative is not simply the product of their individual derivatives. The product rule provides a precise formula for differentiating such expressions, and it is one of the most essential tools in calculus.
当两个函数相乘时，它们的导数并不是各自导数的简单乘积。乘积法则为这类表达式的求导提供了精确的公式，是微积分中最基础也最重要的工具之一。
1. Statement of the Product Rule | 乘积法则的表述
If \(u(x)\) and \(v(x)\) are both differentiable functions of \(x\), then the derivative of their product is given by:
若 \(u(x)\) 和 \(v(x)\) 都是关于 \(x\) 的可导函数，则它们乘积的导数由下式给出：
(uv)′ = u′v + uv′
In words: the derivative of the first function times the second, plus the first function times the derivative of the second.
用语言表述为：第一个函数的导数乘以第二个函数，再加上第一个函数乘以第二个函数的导数。
2. Why Not Just Multiply Derivatives? | 为什么不能直接乘导数？
Consider the simple example \(y = x \cdot x = x^2\). The derivative is \(2x\). If we wrongly multiplied the derivatives, we would get \(1 \cdot 1 = 1\), which is clearly incorrect.
考虑简单例子 \(y = x \cdot x = x^2\)，其导数为 \(2x\)。如果错误地将导数直接相乘，得到 \(1 \cdot 1 = 1\)，这显然不对。
The product rule captures the interaction between the two functions as both change simultaneously. Each function contributes to the rate of change of the whole product.
乘积法则捕捉了两个函数同时变化时相互作用的效应。每个函数都对整体乘积的变化率有所贡献。
3. Derivation from First Principles | 从第一原理推导
Let \(y = uv\), where \(u\) and \(v\) are functions of \(x\). By the definition of the derivative:
设 \(y = uv\)，其中 \(u\) 和 \(v\) 是 \(x\) 的函数。根据导数的定义：
dy/dx = lim (Δx→0) [u(x+Δx)v(x+Δx) − u(x)v(x)] / Δx
Subtract and add \(u(x+Δx)v(x)\) in the numerator, then factor:
在分子中减去再加上 \(u(x+Δx)v(x)\)，然后因式分解：
= lim [u(x+Δx) − u(x)]/Δx · v(x) + u(x+Δx) · [v(x+Δx) − v(x)]/Δx
Taking the limit as Δx → 0 gives \(u′v + uv′\), provided both derivatives exist.
当 Δx → 0 时取极限得到 \(u′v + uv′\)，前提是两个导数都存在。
4. Basic Example: Polynomial and Trigonometric Function | 基础示例：多项式与三角函数
Differentiate \(y = x^2 \sin x\).
求 \(y = x^2 \sin x\) 的导数。
Let \(u = x^2\), so \(u′ = 2x\). Let \(v = \sin x\), so \(v′ = \cos x\).
设 \(u = x^2\)，则 \(u′ = 2x\)。设 \(v = \sin x\)，则 \(v′ = \cos x\)。
dy/dx = 2x · sin x + x² · cos x
This compact form is perfectly acceptable as a final answer.
这个紧凑形式作为最终答案是完全可以接受的。
5. Example with Exponential and Polynomial | 指数函数与多项式的示例
Differentiate \(y = e^x \ln x\).
求 \(y = e^x \ln x\) 的导数。
Here \(u = e^x\), \(u′ = e^x\); \(v = \ln x\), \(v′ = 1/x\).
这里 \(u = e^x\)，\(u′ = e^x\)；\(v = \ln x\)，\(v′ = 1/x\)。
dy/dx = e^x · ln x + e^x · (1/x) = e^x (ln x + 1/x)
Factoring out \(e^x\) simplifies the expression and reduces the risk of errors in later steps.
提取公因式 \(e^x\) 可以简化表达式，并降低后续步骤出错的风险。
6. Special Case: Constant Multiple | 特殊情况：常数倍
If \(v(x) = c\), a constant, then \(v′ = 0\). The product rule becomes:
若 \(v(x) = c\)，即常数，则 \(v′ = 0\)。乘积法则变为：
(cu)′ = c′u + cu′ = 0 · u + c u′ = c u′
This confirms that the constant multiple rule — \((cf)′ = cf′\) — is a special case of the product rule.
这证实了常数倍法则 \((cf)′ = cf′\) 是乘积法则的一个特例。
7. Product of Three Functions | 三个函数相乘的情形
For \(y = uvw\), the product rule extends naturally:
对于 \(y = uvw\)，乘积法则可自然推广：
(uvw)′ = u′vw + uv′w + uvw′
Differentiate \(y = x^2 e^x \cos x\):
求 \(y = x^2 e^x \cos x\) 的导数：
dy/dx = 2x e^x cos x + x² e^x cos x − x² e^x sin x
Each term has exactly one derivative applied, and the other two functions remain unchanged.
每一项中恰好有一个函数被求导，另外两个函数保持不变。
8. Combining Product Rule with Chain Rule | 乘积法则与链式法则的结合
Many exam questions require both rules together. Differentiate \(y = x^2 \sin(3x+1)\).
许多考试题目需要同时使用两个法则。求 \(y = x^2 \sin(3x+1)\) 的导数。
Let \(u = x^2\), \(u′ = 2x\). Let \(v = \sin(3x+1)\). By the chain rule, \(v′ = 3\cos(3x+1)\).
设 \(u = x^2\)，\(u′ = 2x\)。设 \(v = \sin(3x+1)\)。由链式法则，\(v′ = 3\cos(3x+1)\)。
dy/dx = 2x sin(3x+1) + 3x² cos(3x+1)
Always apply the chain rule to the inner function before multiplying by the outer factor.
务必先对内层函数应用链式法则，再乘以外部因式。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One frequent error is writing \(u′v′\) instead of \(u′v + uv′\). Another is forgetting to apply the product rule at all when the expression involves two obvious functions multiplied together.
一个常见错误是写成 \(u′v′\)，而正确应为 \(u′v + uv′\)。另一个错误是当表达式明显为两个函数相乘时，忘记使用乘积法则。
To avoid these, always write down \(u\), \(u′\), \(v\), and \(v′\) separately before combining them. This structured approach minimises slips.
为避免这些错误，务必先分别列出 \(u\)、\(u′\)、\(v\) 和 \(v′\)，再进行组合。这种结构化方法能最大程度地减少失误。
10. Choosing the Right Method | 选择正确的方法
Sometimes an expression can be simplified before differentiation. For instance, \(y = (x+1)(x−1)\) can be expanded to \(x^2−1\), whose derivative is \(2x\). The product rule gives the same result but takes more steps.
有时可以先将表达式化简再求导。例如，\(y = (x+1)(x−1)\) 可展开为 \(x^2−1\)，其导数为 \(2x\)。乘积法则也能得到相同结果，但步骤更多。
However, for expressions like \(x^2 \sin x\) or \(e^x \cos x\), expansion is impossible, and the product rule is the only efficient approach.
然而，对于 \(x^2 \sin x\) 或 \(e^x \cos x\) 这类表达式，无法展开，乘积法则便是唯一高效的求解方法。
11. Practice Problems | 练习题
Differentiate each of the following with respect to \(x\):
对下列函数分别关于 \(x\) 求导：
- \(y = x^3 \ln x\)
- \(y = e^{2x} \tan x\)
- \(y = (x^2+1) \sqrt{x}\)
- \(y = x \cdot 2^x\)
Answers are obtained by setting \(u\) and \(v\) appropriately and applying the product rule systematically.
通过合理设定 \(u\) 和 \(v\) 并系统应用乘积法则，即可得到答案。
12. Summary | 总结
The product rule is a fundamental differentiation technique that must be mastered. It states that \((uv)′ = u′v + uv′\), and it extends to products of three or more functions.
乘积法则是必须掌握的微积分基本技巧。其核心公式为 \((uv)′ = u′v + uv′\)，并可以推广到三个或更多函数相乘的情形。
Practice with a variety of function pairs — polynomial, trigonometric, exponential, and logarithmic — to build fluency and confidence.
通过与不同类型的函数对——多项式、三角函数、指数函数和对数函数——进行组合练习，可以提升熟练度和信心。
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