Advanced Differentiation and Applications | 微分法进阶与应用

📚 Advanced Differentiation and Applications | 微分法进阶与应用

Differentiation is a cornerstone of A-Level Mathematics. Beyond the basic power rule, CIE candidates must master the chain rule, product rule, quotient rule, implicit differentiation, and parametric differentiation — and then apply these tools to tangents, normals, stationary points, rates of change, and optimisation. This guide consolidates the advanced techniques and their most common exam applications in one coherent revision resource.

微分法是 A-Level 数学的核心基石。除了基本幂法则之外，CIE 考生必须熟练掌握链式法则、乘法法则、除法法则、隐函数微分法和参数方程微分法——并运用这些工具处理切线、法线、驻点、变化率与最优化问题。本指南将进阶技巧及其最常见的考试应用整合于一份系统的复习资料中。

1. The Chain Rule | 链式法则

The chain rule differentiates composite functions. If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In function notation, if y = f(u) where u = g(x), then dy/dx = (dy/du) × (du/dx). This rule is indispensable for powers of functions, exponentials, trigonometric functions, and logarithms of non-linear expressions.

链式法则用于对复合函数求导。若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。用函数记号表示，若 y = f(u) 且 u = g(x)，则 dy/dx = (dy/du) × (du/dx)。该法则在处理函数的幂、指数函数、三角函数及非线性表达式的对数时不可或缺。

Common CIE examples include:

CIE 常见范例包括：

y = (3x² + 2)⁵ ⇒ dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴
y = e^(sin x) ⇒ dy/dx = e^(sin x) × cos x
y = ln(2x + 1) ⇒ dy/dx = 2/(2x + 1)
y = sin³(2x) ⇒ dy/dx = 3sin²(2x) × 2cos(2x) = 6sin²(2x)cos(2x)

d/dx [f(g(x))] = f'(g(x)) · g'(x)

The key is to identify the ‘outer function’ and the ‘inner function’, differentiate each, and multiply. In CIE mark schemes, one mark is typically awarded for correctly differentiating the outer function and another for the derivative of the inner function.

关键在于识别“外层函数”与“内层函数”，分别求导后相乘。在 CIE 评分标准中，通常对正确求出外层函数导数和内层函数导数各给一分。

2. The Product Rule | 乘法法则

The product rule handles derivatives of products of two functions. If y = uv, then dy/dx = u(dv/dx) + v(du/dx). A common memory aid is “one times the derivative of the other, plus the other times the derivative of the one.” The rule applies to any two differentiable functions, including polynomials, exponentials, trigonometric functions, and logarithms.

乘法法则用于处理两个函数乘积的导数。若 y = uv，则 dy/dx = u(dv/dx) + v(du/dx)。常用助记口诀是“第一个函数乘第二个函数的导数，加上第二个函数乘第一个函数的导数”。该法则适用于任意两个可导函数，包括多项式、指数函数、三角函数和对数函数。

Worked example: differentiate y = x²e^(3x).

例题：对 y = x²e^(3x) 求导。

Let u = x², v = e^(3x)
du/dx = 2x, dv/dx = 3e^(3x)
dy/dx = x² · 3e^(3x) + e^(3x) · 2x = xe^(3x)(3x + 2)

d/dx (uv) = u dv/dx + v du/dx

Examiners often combine the product rule with the chain rule in a single question, such as y = x·ln(x² + 1), where differentiating ln(x² + 1) itself requires the chain rule.

考官常将乘法法则与链式法则结合在同一道题中，例如 y = x·ln(x² + 1)，其中对 ln(x² + 1) 求导本身就需要链式法则。

3. The Quotient Rule | 除法法则

The quotient rule differentiates ratios of two functions. If y = u/v, then dy/dx = [v(du/dx) − u(dv/dx)] / v². The minus sign in the numerator is a frequent source of sign errors — always differentiate the numerator first, and remember the denominator is squared.

除法法则用于对两个函数之商求导。若 y = u/v，则 dy/dx = [v(du/dx) − u(dv/dx)] / v²。分子中的负号是符号错误的常见来源——务必先对分子求导，并谨记分母要平方。

Worked example: differentiate y = x/(x² + 1).

例题：对 y = x/(x² + 1) 求导。

Let u = x, v = x² + 1
du/dx = 1, dv/dx = 2x
dy/dx = [(x² + 1)(1) − x(2x)] / (x² + 1)² = (1 − x²)/(x² + 1)²

d/dx (u/v) = (v du/dx − u dv/dx) / v²

An alternative is to rewrite u/v as u·v⁻¹ and apply the product rule, which is often faster for simple quotients. However, the quotient rule is safer for complex rational functions and is explicitly listed in the CIE formula booklet.

另一种方法是将 u/v 改写为 u·v⁻¹ 后运用乘法法则，对于简单商式往往更快。但对于复杂的有理函数，除法法则更为稳妥，且 CIE 公式手册中已明确列出。

4. Implicit Differentiation | 隐函数微分法

Implicit differentiation is required when y cannot be written explicitly as a function of x, such as x² + y² = 25 or x³ + y³ = 6xy. Differentiate both sides with respect to x, treating y as a function of x, and apply the chain rule to every term containing y: d/dx (yⁿ) = nyⁿ⁻¹ · dy/dx.

当 y 无法显式写成 x 的函数时，需要用到隐函数微分法，例如 x² + y² = 25 或 x³ + y³ = 6xy。对方程两边分别关于 x 求导，将 y 视为 x 的函数，并对所有含 y 的项运用链式法则：d/dx (yⁿ) = nyⁿ⁻¹ · dy/dx。

Worked example: find dy/dx for x² + y² = 25.

例题：求 x² + y² = 25 的 dy/dx。

Differentiate both sides: 2x + 2y·(dy/dx) = 0
Solve: dy/dx = −x/y

d/dx (yⁿ) = nyⁿ⁻¹ · dy/dx

The derivative of a product like xy requires the product rule: d/dx (xy) = x(dy/dx) + y. After differentiating, collect all dy/dx terms on one side and factorise. Implicit differentiation is essential for finding gradients of circles, ellipses, and other curves not expressible as single explicit functions.

对 xy 这类乘积求导需要乘法法则：d/dx (xy) = x(dy/dx) + y。求导后将所有含 dy/dx 的项移至一侧并提取公因式。隐函数微分法对于求圆、椭圆及其他无法用单一显式函数表示的曲线的斜率至关重要。

5. Parametric Differentiation | 参数方程微分法

When x and y are both given as functions of a parameter t, the gradient is found using dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0. This technique avoids eliminating the parameter and is especially useful for curves defined parametrically, such as x = t², y = t³.

当 x 和 y 均为参数 t 的函数时，切线斜率通过 dy/dx = (dy/dt) ÷ (dx/dt) 求出，前提是 dx/dt ≠ 0。该技巧无需消去参数，特别适用于参数定义的曲线，例如 x = t²，y = t³。

Worked example: x = t², y = t³. Find dy/dx at t = 2.

例题：x = t²，y = t³。求 t = 2 处的 dy/dx。

dx/dt = 2t, dy/dt = 3t²
dy/dx = 3t² / 2t = 3t/2
At t = 2: dy/dx = 3 × 2 ÷ 2 = 3

dy/dx = (dy/dt) ÷ (dx/dt)

For the second derivative, differentiate dy/dx with respect to t and divide by dx/dt again. Note that d²y/dx² ≠ (d²y/dt²) ÷ (d²x/dt²) — this is a classic CIE trap. Always differentiate the first derivative as a function of t, then divide by dx/dt.

对于二阶导数，先将 dy/dx 关于 t 求导，再除以 dx/dt。注意 d²y/dx² ≠ (d²y/dt²) ÷ (d²x/dt²)——这是 CIE 的经典陷阱。务必先将一阶导数视为 t 的函数求导，再除以 dx/dt。

6. Second Derivatives | 二阶导数

The second derivative, d²y/dx², measures the rate of change of the gradient. It determines concavity: when d²y/dx² > 0, the curve is convex (concave upward); when d²y/dx² < 0, the curve is concave downward. It also classifies stationary points: a stationary point with d²y/dx² > 0 is a minimum, and one with d²y/dx² < 0 is a maximum.

二阶导数 d²y/dx² 度量斜率的变化率，用于判定凹凸性：当 d²y/dx² > 0 时，曲线为凸（凹向上）；当 d²y/dx² < 0 时，曲线凹向下。同时它用于分类驻点：d²y/dx² > 0 的驻点为极小值点，d²y/dx² < 0 的驻点为极大值点。

If d²y/dx² = 0 at a stationary point, the test is inconclusive. In such cases, examine the sign of dy/dx just either side of the stationary point. For implicit or parametric curves, compute d²y/dx² carefully using the respective chain rules described above.

若驻点处 d²y/dx² = 0，则该判别法失效。此时需检查驻点两侧 dy/dx 的符号。对于隐函数或参数方程曲线，请使用上述对应的链式法则仔细计算 d²y/dx²。

y” > 0 ⇒ minimum; y” < 0 ⇒ maximum; y'' = 0 ⇒ inconclusive

In CIE questions, you are frequently asked to “find and classify the stationary points” — this requires setting dy/dx = 0, solving for x (or t), and then using d²y/dx² (or a sign table) to classify each point.

在 CIE 试题中，常见要求是“求驻点并分类”——这需要令 dy/dx = 0 解出 x（或 t），再利用 d²y/dx²（或符号表）对每一点进行分类。

7. Tangents and Normals | 切线与法线

The tangent line at a point (x₁, y₁) on a curve has gradient m = dy/dx evaluated at that point. Its equation is y − y₁ = m(x − x₁). The normal line is perpendicular to the tangent, so its gradient is −1/m (provided m ≠ 0), and its equation is y − y₁ = (−1/m)(x − x₁).

曲线上点 (x₁, y₁) 处的切线斜率为该点处 dy/dx 的值 m。切线方程为 y − y₁ = m(x − x₁)。法线垂直于切线，其斜率为 −1/m（前提 m ≠ 0），方程为 y − y₁ = (−1/m)(x − x₁)。

For curves given implicitly or parametrically, first compute dy/dx at the specific point, then apply the same formulae. A common CIE task is to verify that the tangent passes through a given point or to determine where the normal intersects the x-axis or y-axis.

对于隐式或参数方程给出的曲线，先计算指定点处的 dy/dx，再套用同样公式。CIE 常见题型包括验证切线经过某给定点，或求法线与 x 轴、y 轴的交点。

Tangent: y − y₁ = m(x − x₁) Normal: y − y₁ = −(1/m)(x − x₁)

Remember that if m = 0, the tangent is horizontal and the normal is vertical with equation x = x₁. If m is undefined (vertical tangent), the normal is horizontal with equation y = y₁. Always check such edge cases in the exam.

切记：若 m = 0，切线水平，法线为垂直直线 x = x₁；若 m 不存在（切线垂直），则法线水平，方程为 y = y₁。考试时务必检查此类边缘情况。

8. Stationary Points and Optimisation | 驻点与最优化

Stationary points occur where dy/dx = 0. To find them, solve the equation, compute the corresponding y-values, and classify each point as a maximum, minimum, or point of inflection using d²y/dx² or a sign table. A point of inflection with a horizontal tangent occurs when d²y/dx² = 0 and the concavity changes sign.

驻点出现在 dy/dx = 0 处。求解该方程，计算对应的 y 值，并用 d²y/dx² 或符号表将每一点分类为极大值、极小值或拐点。当 d²y/dx² = 0 且凹凸性改变符号时，出现水平切线的拐点。

Optimisation problems ask you to find the maximum or minimum value of a quantity subject to a constraint. The strategy is:

最优化问题要求在约束条件下求某一量的最大值或最小值。解题策略为：

Write down the quantity Q to be maximised or minimised.
Use the constraint to express Q in terms of a single variable.
Differentiate dQ/dx, set it to zero, and solve.
Classify the stationary point and evaluate the optimal value.

写出需要最大化或最小化的量 Q。
利用约束条件将 Q 表示成单一变量的函数。
求 dQ/dx，令其为零并求解。
对驻点分类并计算最优值。

Classic CIE examples include maximising the volume of a box cut from a sheet, minimising the surface area of a cylinder of fixed volume, and maximising the area enclosed by a fixed perimeter. Always state the domain of the variable — for example, length and width must be non-negative — to discard invalid solutions.

CIE 经典例题包括：由矩形纸板剪去四角折成盒子的最大体积、固定体积圆柱的最小表面积、固定周长围成的最大面积。务必说明变量的定义域——例如长度和宽度必须非负——以排除无效解。

9. Rates of Change and Connected Rates | 变化率与相关变化率

Rates of change are applications of the chain rule in context. If y changes with time t and also depends on x, then dy/dt = dy/dx × dx/dt. This idea extends to any chain: for example, dV/dt = dV/dr × dr/dt for a sphere whose radius grows over time.

变化率是链式法则在实际情境中的应用。若 y 随时间 t 变化且同时依赖于 x，则 dy/dt = dy/dx × dx/dt。这一思想可推广至任意链条：例如，对于半径随时间增长的球体，dV/dt = dV/dr × dr/dt。

Worked example: a circle’s radius increases at a constant rate of 2 cm/s. Find the rate of increase of its area when the radius is 5 cm.

例题：圆半径以 2 cm/s 的恒定速率增长。求半径为 5 cm 时面积的增长率。

A = πr² ⇒ dA/dr = 2πr
dr/dt = 2 cm/s
dA/dt = dA/dr × dr/dt = 2πr × 2 = 4πr
When r = 5: dA/dt = 4π × 5 = 20π cm²/s

dy/dt = dy/dx × dx/dt

Connected rates of change problems often involve geometric formulas: volume of a sphere V = (4/3)πr³, surface area of a sphere A = 4πr², and volume of a cone V = (1/3)πr²h. When two quantities in the formula are related, use similar triangles or the given relation to eliminate one variable before differentiating.

相关变化率问题常涉及几何公式：球体积 V = (4/3)πr³、球表面积 A = 4πr²、圆锥体积 V = (1/3)πr²h。当公式中两个量彼此相关时，利用相似三角形或给定关系先消去一个变量再求导。

10. Small Increments and Approximation | 微小增量与近似

Differentiation provides linear approximations. For a small change δx in x, the corresponding change in y is approximately δy ≈ (dy/dx) × δx. This gives the percentage error relationship: percentage change in y ≈ (dy/dx) × (x/y) × percentage change in x.

微分提供线性近似。当 x 发生微小改变 δx 时，y 的相应改变约为 δy ≈ (dy/dx) × δx。由此可得百分比误差关系：y 的百分比变化 ≈ (dy/dx) × (x/y) × x 的百分比变化。

Worked example: estimate the increase in the volume of a cube of side 10 cm when its side length increases by 0.05 cm.

例题：估算边长为 10 cm 的正方体，当其边长增加 0.05 cm 时体积的增量。

V = x³ ⇒ dV/dx = 3x²
δV ≈ dV/dx × δx = 3(10)² × 0.05 = 150 × 0.05 = 7.5 cm³

δy ≈ (dy/dx) × δx

This approximation is valid only for small δx. In CIE questions, you may be asked to estimate values such as cube roots or the change in area of a circle due to a small measurement error. The formula’s linear nature means the approximation improves as δx approaches zero.

该近似仅对微小 δx 有效。在 CIE 试题中，你可能被要求估算立方根等数值，或由测量微小误差引起的圆面积变化。该公式的线性本质意味着 δx 越接近零，近似越精确。

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Advanced Differentiation and Applications | 微分法进阶与应用