📚 Differentiation: Techniques and Applications | 微分：技巧与应用
In Edexcel A-Level Mathematics, differentiation is one of the most heavily assessed topics across Pure Mathematics, Mechanics and even parts of Statistics. This article reviews the core techniques of differentiation, including first principles, standard rules, chain, product and quotient rules, implicit and parametric differentiation, second derivatives, tangents, normals and optimisation. Understanding these methods is essential not only for Paper 1 and Paper 2 but also for connected rates of change in Mechanics.
在 Edexcel A-Level 数学中，微分是纯数学、力学甚至统计学部分中最常考查的主题之一。本文回顾微分的核心方法，包括第一性原理、基本法则、链式法则、乘积法则与商法则、隐函数与参数方程微分、二阶导数、切线与法线以及优化问题。掌握这些方法不仅对 Paper 1 和 Paper 2 至关重要，也是力学中相关变化率问题的基础。
1. The Definition and First Principles | 定义与第一性原理
The derivative of a function f(x) at a point x is defined as the limit of the average rate of change as the increment h tends to zero.
函数 f(x) 在点 x 处的导数定义为当增量 h 趋近于零时平均变化率的极限。
f'(x) = limₕ→0 [f(x+h) − f(x)] / h
For example, if f(x) = x², then f(x+h) = (x+h)² = x² + 2xh + h². Subtracting f(x) gives 2xh + h², and dividing by h yields 2x + h. As h approaches 0, the derivative is 2x.
例如，若 f(x) = x²，则 f(x+h) = (x+h)² = x² + 2xh + h²。减去 f(x) 得到 2xh + h²，再除以 h 得到 2x + h。当 h 趋近于 0 时，导数为 2x。
Exam questions occasionally ask candidates to prove this result from first principles. You must show the full limit process, not just quote the standard rule.
考试题目偶尔会要求考生从第一性原理证明该结果。你必须展示完整的极限过程，而不能直接引用标准法则。
2. Basic Differentiation Rules | 基本求导法则
For a power function f(x) = xⁿ, where n is a real constant, the derivative is given by:
对于幂函数 f(x) = xⁿ，其中 n 为实常数，其导数由以下公式给出：
d/dx (xⁿ) = nxⁿ⁻¹
This rule also applies to negative and fractional powers, so expressions such as 1/x and √x should be rewritten as x⁻¹ and x¹ᐟ² before differentiating.
该法则同样适用于负指数和分数指数，因此像 1/x 和 √x 这样的表达式应先改写为 x⁻¹ 和 x¹ᐟ² 再求导。
If a function is multiplied by a constant, the constant is preserved: d/dx [k f(x)] = k f'(x). Also, differentiation is linear, so d/dx [f(x) ± g(x)] = f'(x) ± g'(x).
如果函数乘以常数，则常数保留：d/dx [k f(x)] = k f'(x)。此外，微分是线性的，因此 d/dx [f(x) ± g(x)] = f'(x) ± g'(x)。
For example, d/dx (5x⁴ − 3x² + 2x − 7) = 20x³ − 6x + 2. The constant term −7 differentiates to 0.
例如，d/dx (5x⁴ − 3x² + 2x − 7) = 20x³ − 6x + 2。常数项 −7 的导数为 0。
3. The Product Rule | 乘积法则
When differentiating a product of two functions, y = uv, the derivative is given by:
当对两个函数的乘积 y = uv 求导时，导数由以下公式给出：
dy/dx = u dv/dx + v du/dx
In words: differentiate the first function, keep the second; then keep the first, differentiate the second, and add the two results.
简单来说：先对第一个函数求导并保持第二个函数不变；再保持第一个函数不变并对第二个函数求导，然后将两项相加。
Example: If y = x² sin x, let u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x, so dy/dx = 2x sin x + x² cos x.
示例：若 y = x² sin x，令 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x，所以 dy/dx = 2x sin x + x² cos x。
The product rule is often required when both factors are non-constant, such as x eˣ or x ln x. Always identify u and v clearly before applying the formula.
当两个因子都是非常数时，例如 x eˣ 或 x ln x，通常需要使用乘积法则。在应用公式前务必明确 u 和 v。
4. The Quotient Rule | 商法则
For a quotient y = u/v, where u and v are functions of x, the derivative is:
对于商式 y = u/v，其中 u 和 v 都是 x 的函数，其导数为：
dy/dx = (v du/dx − u dv/dx) / v²
The denominator is always the square of the original denominator. It is useful to remember the mnemonic “low d high minus high d low over low squared”.
分母始终是原分母的平方。记住口诀“下乘上导减去上乘下导，再除以下平方”会很有帮助。
Example: If y = x / (x² + 1), let u = x and v = x² + 1. Then dy/dx = [(x²+1)(1) − x(2x)] / (x²+1)² = (1 − x²) / (x²+1)².
示例：若 y = x / (x² + 1)，令 u = x，v = x² + 1。则 dy/dx = [(x²+1)(1) − x(2x)] / (x²+1)² = (1 − x²) / (x²+1)²。
Be careful with the subtraction order in the numerator. A common error is writing u dv/dx − v du/dx, which gives the wrong sign.
注意分子中的减法顺序。一个常见错误是写成 u dv/dx − v du/dx，这会导致符号错误。
5. The Chain Rule | 链式法则
The chain rule is used to differentiate composite functions. If y is a function of u and u is a function of x, then:
链式法则用于对复合函数求导。如果 y 是 u 的函数，u 是 x 的函数，则：
dy/dx = dy/du × du/dx
For a simple composite function such as y = (3x² + 2)⁵, let u = 3x² + 2. Then y = u⁵, so dy/du = 5u⁴ and du/dx = 6x. Hence dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴.
对于简单的复合函数如 y = (3x² + 2)⁵，令 u = 3x² + 2。则 y = u⁵，因此 dy/du = 5u⁴，du/dx = 6x。所以 dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴。
The chain rule can be extended to functions involving exponentials, logarithms and trigonometric functions, such as eˢⁱⁿˣ or ln(cos x).
链式法则可推广到涉及指数、对数和三角函数的复合函数，例如 eˢⁱⁿˣ 或 ln(cos x)。
In A-Level exams, you are expected to apply the chain rule fluently without always writing out the substitution u, but showing intermediate steps can prevent mistakes.
在 A-Level 考试中，希望你能够熟练应用链式法则，不必总是写出替换变量 u，但展示中间步骤可以防止出错。
6. Derivatives of Exponential and Logarithmic Functions | 指数函数与对数函数的导数
The exponential function eˣ has the special property that its derivative is itself:
指数函数 eˣ 具有特殊的性质，其导数等于它本身：
d/dx (eˣ) = eˣ
More generally, for a constant base a > 0, d/dx (aˣ) = aˣ ln a. For the natural logarithm, d/dx (ln x) = 1/x, for x > 0.
更一般地，对于常数底数 a > 0，d/dx (aˣ) = aˣ ln a。对于自然对数，d/dx (ln x) = 1/x，其中 x > 0。
Using the chain rule, if y = eᶠ⁽ˣ⁾, then dy/dx = f'(x) eᶠ⁽ˣ⁾. Similarly, d/dx [ln(g(x))] = g'(x) / g(x).
利用链式法则，若 y = eᶠ⁽ˣ⁾，则 dy/dx = f'(x) eᶠ⁽ˣ⁾。类似地，d/dx [ln(g(x))] = g'(x) / g(x)。
Example: If y = ln(x² + 3), then dy/dx = 2x / (x² + 3). These exponential and logarithmic rules frequently appear in modelling and growth problems.
示例：若 y = ln(x² + 3)，则 dy/dx = 2x / (x² + 3)。这些指数与对数法则常出现在建模和增长问题中。
7. Derivatives of Trigonometric Functions | 三角函数的导数
The standard derivatives of the three main trigonometric functions are:
三个主要三角函数的标准导数如下：
d/dx (sin x) = cos x
d/dx (cos x) = −sin x
d/dx (tan x) = sec² x
These results must be memorised and used with the chain rule when the angle is a function of x. For example, d/dx [sin(2x)] = 2 cos(2x).
这些结果必须牢记，并且当角度是 x 的函数时需要与链式法则结合使用。例如，d/dx [sin(2x)] = 2 cos(2x)。
The derivatives of reciprocal functions such as sec x, cosec x and cot x are also in the Edexcel specification. In particular, d/dx (sec x) = sec x tan x, d/dx (cosec x) = −cosec x cot x and d/dx (cot x) = −cosec² x.
倒数函数的导数，如 sec x、cosec x 和 cot x 也在 Edexcel 考试范围内。特别是 d/dx (sec x) = sec x tan x，d/dx (cosec x) = −cosec x cot x，d/dx (cot x) = −cosec² x。
8. Implicit Differentiation | 隐函数微分
When a curve is given by an equation involving both x and y, such as x² + y² = 25, we can differentiate both sides with respect to x. Remember that y is a function of x, so differentiating y² requires the chain rule.
当曲线由同时包含 x 和 y 的方程给出时，例如 x² + y² = 25，我们可以对方程两边关于 x 求导。记住 y 是 x 的函数，因此对 y² 求导需要使用链式法则。
d/dx (x² + y²) = d/dx (25) → 2x + 2y (dy/dx) = 0
Solving gives dy/dx = −x/y. The process of finding dy/dx without explicitly solving for y is called implicit differentiation.
求解可得 dy/dx = −x/y。这种不求 y 的显式表达式而直接求 dy/dx 的过程称为隐函数微分。
In general, a term yⁿ differentiates to n yⁿ⁻¹ dy/dx. This method is essential for curves such as circles, ellipses and other relations that are not functions in explicit form.
一般来说，项 yⁿ 的导数为 n yⁿ⁻¹ dy/dx。该方法对于圆、椭圆以及其他非显式函数关系的曲线至关重要。
9. Parametric Differentiation | 参数方程微分
If a curve is defined parametrically by x = f(t) and y = g(t), then the derivative dy/dx can be found from:
如果曲线由参数方程 x = f(t) 和 y = g(t) 定义，则导数 dy/dx 可以通过以下公式求得：
dy/dx = (dy/dt) / (dx/dt)
This formula comes from the chain rule. It is valid provided dx/dt ≠ 0. You should find dy/dt and dx/dt separately before dividing.
该公式来源于链式法则。只要 dx/dt ≠ 0 即可使用。在相除之前，应分别求出 dy/dt 和 dx/dt。
Example: If x = t² and y = 2t + 1, then dx/dt = 2t and dy/dt = 2, so dy/dx = 2 / 2t = 1/t.
示例：若 x = t²，y = 2t + 1，则 dx/dt = 2t，dy/dt = 2，因此 dy/dx = 2 / 2t = 1/t。
For the second derivative with respect to x, use d²y/dx² = d/dt (dy/dx) / (dx/dt). This is often tested in questions on tangents, normals or stationary points of parametric curves.
对于关于 x 的二阶导数，使用 d²y/dx² = d/dt (dy/dx) / (dx/dt)。这在参数曲线的切线、法线或驻点问题中经常考查。
10. Second Derivatives and Classifying Stationary Points | 二阶导数与驻点分类
The second derivative f”(x) is the derivative of f'(x). It represents the rate of change of the gradient and helps determine the concavity of a curve.
二阶导数 f”(x) 是 f'(x) 的导数。它表示斜率的变化率，有助于确定曲线的凹凸性。
At a stationary point, f'(x) = 0. To classify it, evaluate f”(x) at that point: if f”(x) > 0, the point is a local minimum; if f”(x) < 0, it is a local maximum; if f''(x) = 0, the test is inconclusive and you should check the sign of f'(x) on either side.
在驻点处，f'(x) = 0。为了分类该驻点，计算该点的 f”(x)：若 f”(x) > 0，该点是局部极小值；若 f”(x) < 0，该点是局部极大值；若 f''(x) = 0，检验失效，应检查 f'(x) 在该点两侧的符号。
Example: For y = x³ − 3x, f'(x) = 3x² − 3 and f”(x) = 6x. Setting f'(x) = 0 gives x = ±1. At x = 1, f”(1) = 6 > 0, so a local minimum; at x = −1, f”(−1) = −6 < 0, so a local maximum.
示例：对于 y = x³ − 3x，f'(x) = 3x² − 3，f”(x) = 6x。令 f'(x) = 0 得 x = ±1。在 x = 1 处，f”(1) = 6 > 0，为局部极小值；在 x = −1 处，f”(−1) = −6 < 0，为局部极大值。
11. Tangents, Normals and Optimisation | 切线、法线与优化
The derivative at a point gives the gradient of the tangent to the curve at that point. The equation of the tangent at (x₁, y₁) is:
曲线在某点的导数给出了该点切线的斜率。在 (x₁, y₁) 处的切线方程为：
y − y₁ = m (x − x₁), where m = dy/dx at (x₁, y₁)
The normal is perpendicular to the tangent, so its gradient is −1/m. The normal equation is y − y₁ = (−1/m)(x − x₁).
法线垂直于切线，因此其斜率为 −1/m。法线方程为 y − y₁ = (−1/m)(x − x₁)。
Optimisation problems involve finding the maximum or minimum of a quantity such as area, volume, surface area, distance or cost. You must set up an expression for the quantity, differentiate it with respect to one variable, set the derivative equal to zero and justify the nature of the stationary point.
优化问题涉及求某个量的最大值或最小值，例如面积、体积、表面积、距离或成本。你必须建立该量的表达式，对其中一个变量求导，令导数等于零，并说明驻点的性质。
Example: A rectangular box with a square base and no top has volume 500 cm³. If the base side length is x, find the value of x that minimises the surface area. The first step is to express the height h = 500 / x², then write the surface area A = x² + 4xh = x² + 2000 / x.
示例：一个无盖长方体盒子的底面为正方形，体积为 500 cm³。若底面边长为 x，求使表面积最小的 x 值。第一步是表达高度 h = 500 / x²，然后写出表面积 A = x² + 4xh = x² + 2000 / x。
Differentiating gives dA/dx = 2x − 2000 / x². Setting this to zero gives 2x³ = 2000, so x³ = 1000 and x = 10. The second derivative d²A/dx² = 2 + 4000 / x³ is positive at x = 10, confirming a minimum.
求导得 dA/dx = 2x − 2000 / x²。令其为零得 2x³ = 2000，所以 x³ = 1000，x = 10。二阶导数 d²A/dx² = 2 + 4000 / x³ 在 x = 10 处为正，确认是最小值。
12. Common Mistakes and Exam Tips | 常见错误与考试技巧
One common mistake is forgetting to rewrite terms like 1/x, √x or 1/x² as powers before differentiating. This leads to errors when applying the power rule.
一个常见错误是在求导前忘记将 1/x、√x 或 1/x² 等项改写成幂的形式。这会导致应用幂法则时出错。
Another frequent error is applying the product or quotient rule when one factor is constant, which is unnecessary. For example, d/dx (5x²) can be done directly as 10x rather than using the product rule.
另一个常见错误是在其中一个因子为常数时仍然使用乘积法则或商法则，这是不必要的。例如，d/dx (5x²) 可直接求导为 10x，而不必使用乘积法则。
In implicit differentiation, candidates often forget to multiply by dy/dx when differentiating a y-term. Always write d/dx (yⁿ) = n yⁿ⁻¹ dy/dx.
在隐函数微分中，考生常忘记对 y 的项求导时乘以 dy/dx。务必写出 d/dx (yⁿ) = n yⁿ⁻¹ dy/dx。
Finally, when classifying stationary points, do not rely solely on the second derivative if f”(x) = 0. In such cases, use a sign table for f'(x) around the stationary point to determine whether it is a maximum, minimum or point of inflection.
最后，在分类驻点时，如果 f”(x) = 0，不要只依赖二阶导数。此时应使用 f'(x) 在驻点附近的符号表来判断该点是极大值、极小值还是拐点。
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