Investigation 2 - Differentiating Composite Functions | 探究2：复合函数的微分

📚 Investigation 2 - Differentiating Composite Functions | 探究2：复合函数的微分

In IB Mathematics, you encounter many functions that are built by nesting one function inside another. Differentiating such composite functions requires a powerful technique known as the Chain Rule. This investigation will guide you through understanding what composite functions are, why the Chain Rule works, and how to apply it in a variety of contexts, from simple polynomials to trigonometric, exponential, and logarithmic functions. Mastering this rule is essential for success in calculus and forms the foundation for more advanced topics like implicit differentiation and integration by substitution.

在IB数学中，你会遇到许多由多个函数嵌套而成的函数。对这些复合函数进行微分需要使用一种称为链式法则的强大技巧。本次探究将引导你理解什么是复合函数、链式法则为何有效，以及如何在从简单多项式到三角函数、指数函数和对数函数等多种情境下应用该法则。掌握链式法则对于微积分的成功学习至关重要，并为隐函数微分和换元积分法等更高级的主题打下基础。

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1. Introduction: What Are Composite Functions? | 引言：什么是复合函数？

the result of another function. If we have two functions f and g, the composite function f ∘ g is defined as (f ∘ g)(x) = f(g(x)). We often read this as 'f of g of x' The function g is the inner function, and f is the outer function. For example, if f(x) = √x and g(x) = x² + 1, then the composite function h(x) = √x² + 1 is formed by taking g(x) first and then applying f. Recognising the inner and outer parts of a composite function is the first critical step in differentiation.

当一个函数作用于另一个函数的结果时，就产生了复合函数。如果我们有两个函数 f 和 g，复合函数 f ∘ g 定义为 (f ∘ g)(x) = f(g(x))，通常读作“f of g of x”。这里 g 是内层函数，f 是外层函数。例如，若 f(x) = √x 而 g(x) = x² + 1，则复合函数 h(x) = √x² + 1 就是先取 g(x) 再应用 f 得到的。识别复合函数的内层和外层部分是微分的第一步，也是最关键的一步。

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2. The Chain Rule – Intuition and Statement | 链式法则——直观与陈述

with respect to x. This insight leads to the Chain Rule. Formally, if y = f(u) and u = g(x) are differentiable functions, then y is a differentiable function of x, and the derivative is given by dy/dx = dy/du × du/dx. In function notation, this is often written as: if h(x) = f(g(x)), then h′(x) = f′(g(x)) · g′(x).

设想你正在绘制 y = (3x + 2)⁴ 的图像。x 的微小变化会引起内层函数 u = 3x + 2 的变化，进而引起外层函数 y = u⁴ 的变化。因此，y 对 x 的变化率等于 y 对 u 的变化率与 u 对 x 的变化率的乘积。这一洞见便引出了链式法则。正式表述为：若 y = f(u) 且 u = g(x) 都是可微函数，则 y 也是 x 的可微函数，导数为 dy/dx = dy/du × du/dx。用函数符号表示，常写作：若 h(x) = f(g(x))，则 h′(x) = f′(g(x)) · g′(x)。

dy/dx = f′(g(x)) · g′(x)

This rule tells us to differentiate the outer function, evaluated at the inner function, and then multiply by the derivative of the inner function. It is arguably the most frequently used differentiation rule in calculus.

这条法则告诉我们，先对外层函数求导（在内层函数的值处计算），再乘以内层函数的导数。这可以说是微积分中使用最频繁的微分法则。

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3. Deriving the Chain Rule (Optional Investigation) | 推导链式法则（可选探究）

× k/h, provided k ≠ 0. As h → 0, k → 0 (since g is continuous), so the first factor tends to f′(u) and the second factor tends to g′(x). Hence the derivative is f′(g(x)) · g′(x). This intuitive proof helps explain why the product of rates appears.

虽然并非所有IB课程都要求，但看一看推导过程可以加深理解。考虑 y = f(g(x)) 的差商。令 k = g(x + h) − g(x) 且 u = g(x)。那么 f(g(x + h)) − f(g(x)) = f(u + k) − f(u)。导数就是 h → 0 时的极限。我们可以将 [f(u + k) − f(u)]/h 写成 [f(u + k) − f(u)]/k × k/h，前提是 k ≠ 0。当 h → 0 时，k → 0（因为 g 连续），所以第一个因子趋于 f′(u)，第二个因子趋于 g′(x)。因此导数为 f′(g(x)) · g′(x)。这个直观证明有助于解释为什么会出现变化率的乘积。

/dx = dy/du × du/dx makes the cancellation idea especially clear, although we treat the du's as differentials rather than just cancelling numbers.

莱布尼兹记号 dy/dx = dy/du × du/dx 让这种“约分”的想法格外清晰，尽管我们是将 du 视作微分处理，而非简单的数字约去。

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4. Basic Examples of the Chain Rule | 链式法则的基本例题

+ 5. Then f′(u) = 4u³ and g′(x) = 3. Using the rule, h′(x) = 4(3x + 5)³ × 3 = 12(3x + 5)³.

下面我们将链式法则应用于简单的多项式复合函数。考虑 h(x) = (3x + 5)⁴。找出外层函数 f(u) = u⁴，内层函数 u = g(x) = 3x + 5。则 f′(u) = 4u³，g′(x) = 3。运用链式法则，h′(x) = 4(3x + 5)³ × 3 = 12(3x + 5)³。

5(x² + 2x)⁴ (2x + 2). This can be left as a product or simplified to 10(x + 1)(x² + 2x)⁴, but the factored form is usually acceptable.

第二个例子，对 y = (x² + 2x)⁵ 进行微分。令 u = x² + 2x，则 y = u⁵。dy/du = 5u⁴，du/dx = 2x + 2。因此 dy/dx = 5(x² + 2x)⁴ (2x + 2)。可以保留乘积形式，或者简化为 10(x + 1)(x² + 2x)⁴，但通常保留因式分解后的形式即可。

In general, for any power function (u(x))ⁿ, the derivative is n(u(x))ⁿ⁻¹ · u′(x).

一般地，对于幂函数 (u(x))ⁿ，其导数为 n(u(x))ⁿ⁻¹ · u′(x)。

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5. The Chain Rule with Trigonometric Functions | 链式法则与三角函数

example, differentiate y = sin(2x). Here u = 2x, u′ = 2. So y′ = cos(2x) · 2 = 2 cos(2x).

三角函数经常与线性或多项式自变量一起出现，因此构成了复合函数。sin(u) 的导数是 cos(u) · u′，其他三角函数类似。例如，对 y = sin(2x) 微分。这里 u = 2x，u′ = 2。因此 y′ = cos(2x) · 2 = 2 cos(2x)。

Let' with u′ = 3x². The derivative of cos(u) is −sin(u), so y′ = −sin(x³) · 3x² = −3x² sin(x³).

再试一例 y = cos(x³)。内层函数为 u = x³，u′ = 3x²。cos(u) 的导数是 −sin(u)，所以 y′ = −sin(x³) · 3x² = −3x² sin(x³)。

patterns are straightforward once you remember to multiply by the derivative of the angle.

对于正切函数，d/dx[tan(u)] = sec²(u) · u′。因此，若 y = tan(5x + 1)，则 y′ = sec²(5x + 1) · 5 = 5 sec²(5x + 1)。只要记住乘以角度部分的导数，这些模式就非常直接了。

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6. The Chain Rule with Exponential and Logarithmic Functions | 链式法则与指数函数和对数函数

The natural exponential function is particularly elegant. The derivative of e^u is e^u · u′. For instance, differentiate y = e^{2x + 3}. Here u = 2x + 3, u′ = 2, so y′ = e^{2x + 3} · 2 = 2 e^{2x + 3}. The exponential function essentially reproduces itself, but the chain rule still requires the inner derivative.

自然指数函数格外优雅。e^u 的导数是 e^u · u′。例如，对 y = e^{2x + 3} 微分。这里 u = 2x + 3，u′ = 2，因此 y′ = e^{2x + 3} · 2 = 2 e^{2x + 3}。指数函数本质上再现了自身，但链式法则仍然需要乘以内层导数。

= g(x), then dy/du = 1/u and du/dx = g′(x), giving dy/dx = (1/g(x)) · g′(x) = g′(x)/g(x).

对数微分同样依赖于链式法则。对于 y = ln(g(x))，导数为 g′(x)/g(x)，前提是 g(x) > 0。这是因为若 y = ln(u) 且 u = g(x)，则 dy/du = 1/u，du/dx = g′(x)，从而 dy/dx = (1/g(x)) · g′(x) = g′(x)/g(x)。

e: If y = ln(x² + 1), then y′ = (2x)/(x² + 1). Example with general exponential: For y = a^u, where a > 0, we rewrite a^u = e^{u ln a}. Then y′ = e^{u ln a} · u′ ln a = a^u · u′ ln a.

示例：若 y = ln(x² + 1)，则 y′ = (2x)/(x² + 1)。一般指数函数的例子：对于 y = a^u（a > 0），我们将其改写为 a^u = e^{u ln a}，则 y′ = e^{u ln a} · u′ ln a = a^u · u′ ln a。

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7. Repeated Chain Rule: Multiple Layers | 重复链式法则：多层函数

Sometimes a function is composed of more than two functions. For example, h(x) = sin(e^x²). Here we have three layers: outermost is sin( ), middle is e fferentiate from outside in, multiplying the derivatives at each step. Write u = x², v = e^u, y = sin(v). Then dy/dx = dy/dv × dv/du × du/dx = cos(v) · e^u · 2x = cos(e^x²) · e^x² · 2x.

有时一个函数由两个以上的函数复合而成。例如，h(x) = sin(e^x²)。这里有三个层次：最外层是 sin( )，中间层是 e ，最内层是 x²。链式法则可以自然推广：从外向内逐层求导，每一步都乘上相应层的导数。令 u = x²，v = e^u，y = sin(v)。则 dy/dx = dy/dv × dv/du × du/dx = cos(v) · e^u · 2x = cos(e^x²) · e^x² · 2x。

The procedure is often described as 'derivative of outer, keep inner intact, times derivative of next inner, and so on'. With practice, you can do this without explicitly substituting variables, but keeping track of the layers is vital for accuracy.

这个过程常被描述为“外层导数保持内层不变，乘以下一层的导数，依此类推”。经过练习，你可以无需显式变量替换即完成求导，但随时追踪各个层次对准确性至关重要。

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8. Applications: Tangent and Normal Lines | 应用：切线与法线

n point often requires the derivative, which may involve the Chain Rule. For the curve y = √x² + 9, find the tangent at x = 4. First, rewrite y = (x² + 9)^½. The derivative is y′ = ½(x² + 9)^−½ · 2x = x / √x² + 9. At x = 4, y = √16 + 9 ts equation is y − 5 = (4/5)(x − 4). The normal line is perpendicular, with gradient −5/4.

求曲线在某点的切线方程经常需要用到导数，这可能涉及链式法则。对于曲线 y = √x² + 9，求 x = 4 处的切线。首先，将 y 重写为 y = (x² + 9)^½。导数为 y′ = ½(x² + 9)^−½ · 2x = x / √x² + 9。当 x = 4 时，y = √16 + 9 = 5，且 y′ = 4/5。因此切线斜率为 4/5，经过点 (4,5)。其方程为 y − 5 = (4/5)(x − 4)。法线与切线垂直，斜率为 −5/4。

Such problems illustrate how the Chain Rule seamlessly provides gradients for curves defined by composite expressions. It's worth practicing setting up the derivative expression carefully before plugging in values.

这类问题展示了链式法则如何无缝地为复合表达式定义的曲线提供斜率。值得在代入数值前先仔细列出导数表达式并多加练习。

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9. Using the Chain Rule in Related Rates (Optional) | 链式法则在相关变化率中的应用（可选）

radius r of a circle increases at 3 cm/s. How fast is the area A increasing when r = 10 cm? dt = 3, so dA/dt = 2π(10)(3) = 60π cm²/s. The Chain Rule appears as dA/dt = dA/dr × dr/dt.

相关变化率问题是链式法则的经典应用，其中两个或更多量随时间变化。假设一个圆的半径 r 以 3 cm/s 的速度增加。当 r = 10 cm 时，面积 A 的增加速度是多少？已知 A = πr²。两边同时对时间 t 求导，得到 dA/dt = 2πr · dr/dt。当 r = 10，dr/dt = 3，因此 dA/dt = 2π(10)(3) = 60π cm²/s。链式法则体现为 dA/dt = dA/dr × dr/dt。

ch squared term: 2x dx/dt + 2y dy/dt = 0. This allows us to find one rate given the other.

另一个常见例子涉及靠墙滑落的梯子，其高度 y 和离墙的距离 x 满足 x² + y² = L²。对 t 进行隐函数求导时，每个平方项都用到链式法则：2x dx/dt + 2y dy/dt = 0。这使我们能够根据一个变化率求出另一个变化率。

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10. Common Mistakes and Tips | 常见错误与技巧

Even experienced students slip up with the Chain Rule. The most frequent error is forgetting to multiply by the derivative of the inner function. For example, thinking d/dx(sin(x²)) is cos(x²) without the factor 2x. Always ask yourself: 'Have I multiplied by the inner derivative?' Another pitfall is misidentifying the inner function, especially in compositions like ln(sin x) or e^{tan x}.

即便是有经验的学生也会在链式法则上犯错。最常见的错误是忘记乘以内层函数的导数。例如，认为 d/dx(sin(x²)) 就是 cos(x²) 而遗漏因子 2x。一定要问自己：“我乘以内层导数了吗？”另一个陷阱是错误识别内层函数，特别是在 ln(sin x) 或 e^{tan x} 这样的复合中。

When dealing with negative powers or rational exponents, rewrite the function clearly: 1/(2x + 1)ⁿ is (2x + 1)^-n, and then apply the power chain rule. Also, be cautious with constants – the derivative of e³ is zero, while the derivative of e^3x is 3e^3x.

在处理负指数或有理指数时，应明确重写函数：1/(2x + 1)ⁿ 就是 (2x + 1)^-n，然后应用幂链式法则。此外，要留意常数——e³ 的导数为零，而 e^3x 的导数为 3e^3x。

Finally, practice writing intermediate steps. Using the u-substitution explicitly on paper can prevent errors until you build mental fluency. Many IB questions award method marks even if the final answer has a small slip.

最后，练习写出中间步骤。在纸上明确使用 u 代换可以防止出错，直到你达到心算流畅的程度。许多 IB 题目即使最终答案有小纰漏，也会给方法分。

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11. Practice Problems | 练习题

Test your understanding with these problems. Differentiate the following with respect to x:

用以下题目检验你的理解。对 x 求下列函数的导数：

• y = (5x - 2)⁷
• y = √4x + 1
• y = sin(3x² - x)
• y = cos²(x) (Hint: rewrite as (cos x)²)
• y = e^{tan x}
• y = ln(x³ + 3x)
• y = (2x + 1)⁴ (x - 3)³ (requires product rule and chain rule)

Try finding the equation of the tangent to the curve y = √x² + 16 your answers by differentiating step-by-step and looking for the multiplier from the inner function.

尝试求曲线 y = √x² + 16 在 x = 3 处的切线方程。另外，如果球的体积 V = (4/3)πr³ 且半径以 2 cm/min 增长，求 r = 6 cm 时的 dV/dt。通过逐步求导并检查来自内层函数的乘数来验证你的答案。

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12. Summary and Conclusion | 总结与结论

The Chain Rule is an indispensable tool for differentiating composite functions. By breaking a function into its outer and inner layers, then multiplying the respective derivatives, you can handle a vast range of expressions efficiently. From simple polynomials and trig functions to exponentials, logarithms, and even related rates, the same principle applies. Mastery comes with consistent practice and developing the reflex to identify the structure of any function before differentiating. Remember: differentiate the outer, keep the inside, multiply by the derivative of the inside. With this rule firmly in your toolkit, you are well-prepared for the challenges of IB calculus.

链式法则是微分复合函数不可或缺的工具。通过将函数分解为外层和内层，然后将相应的导数相乘，你可以高效处理各种表达式。从简单的多项式和三角函数到指数、对数，乃至相关变化率，这一原理都始终适用。掌握链式法则需要持续练习，并养成在求导前先识别函数结构的习惯。记住：对外层求导，保持内层，再乘以内层的导数。将这条法则稳固地纳入你的工具箱后，你就能从容应对 IB 微积分中的挑战。

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