📚 Differentiation Techniques and Applications | Edexcel A-Level 纯数学复习：微分方法与应用
In Edexcel A-Level Mathematics, differentiation appears in Pure Mathematics papers and underpins many modelling and problem-solving questions. This revision guide covers the essential derivative rules, standard functions, parametric and implicit methods, second derivatives, and common applications such as tangents, stationary points and rates of change.
在 Edexcel A-Level 数学中，微分贯穿纯数学试卷，并支撑许多建模与解题问题。本复习指南涵盖基本求导法则、常见函数、参数方程与隐函数求导、二阶导数，以及切线、驻点和变化率等常见应用。
1. Differentiation from First Principles | 导数第一性原理
The derivative is defined using a limiting process. It gives the gradient of the tangent to the curve at a particular point.
导数通过极限过程定义。它给出了曲线上某一点处切线的斜率。
f'(x) = limₕ→₀ [f(x+h) − f(x)] / h
This limiting form is frequently tested in Edexcel Pure Mathematics. You may be asked to prove the derivative of a simple polynomial function from first principles.
这种极限形式在 Edexcel 纯数学中经常考查。你可能需要根据第一性原理证明简单多项式函数的导数。
For example, if f(x) = x², then f'(x) = limₕ→₀ [(x+h)² − x²] / h = limₕ→₀ (2xh + h²) / h = limₕ→₀ (2x + h) = 2x.
例如，若 f(x) = x²，则 f'(x) = limₕ→₀ [(x+h)² − x²] / h = limₕ→₀ (2xh + h²) / h = limₕ→₀ (2x + h) = 2x。
2. Basic Rules of Differentiation | 基本微分法则
The power rule is the most basic and frequently used differentiation rule. It states that the derivative of x to the power n is n times x to the power n minus one.
幂法则是求导中最基本、最常用的法则。它指出 x 的 n 次幂的导数等于 n 乘以 x 的 n 减一次幂。
d/dx [xⁿ] = n xⁿ⁻¹
Two further basic results are the constant multiple rule and the sum rule. These allow the derivative of a polynomial term by term.
另外两个基本法则是常数倍法则和加法法则。它们允许对多项式逐项求导。
For instance, d/dx (5x³ − 2x + 7) = 15x² − 2. The constant term 7 differentiates to zero because a horizontal line has zero gradient.
例如，d/dx (5x³ − 2x + 7) = 15x² − 2。常数项 7 的导数为零，因为水平线的斜率为零。
3. Chain Rule | 链式法则
The chain rule is used when one function is nested inside another. It allows the derivative of a composite function such as y = (3x² + 5)⁴ to be found efficiently.
当一个函数嵌套在另一个函数中时使用链式法则。它可以高效地求出复合函数的导数，例如 y = (3x² + 5)⁴。
dy/dx = dy/du × du/dx
Let u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x. Multiplying gives dy/dx = 4u³ × 6x = 24x(3x² + 5)³.
设 u = 3x² + 5，则 y = u⁴。因此 dy/du = 4u³，du/dx = 6x。相乘得到 dy/dx = 4u³ × 6x = 24x(3x² + 5)³。
In Edexcel questions, the chain rule is often combined with other rules, so always identify the outer and inner functions clearly before differentiating.
在 Edexcel 考题中，链式法则常与其他法则结合使用，因此求导前务必明确区分外层函数和内层函数。
4. Product Rule | 乘积法则
When two functions are multiplied together, the product rule is required. It is not correct simply to multiply the individual derivatives.
当两个函数相乘时，需要使用乘积法则。仅仅将两个函数各自的导数相乘是不正确的。
d/dx [u v] = u’v + u v’
For y = x² eˣ, let u = x² and v = eˣ. Then u’ = 2x and v’ = eˣ, so dy/dx = 2x eˣ + x² eˣ = x eˣ(2 + x).
对于 y = x² eˣ，设 u = x²，v = eˣ。则 u’ = 2x，v’ = eˣ，因此 dy/dx = 2x eˣ + x² eˣ = x eˣ(2 + x)。
Always write both terms and then simplify. A common error is to miss the derivative of the first or second function.
务必写出两项再进行化简。常见错误是漏掉第一个或第二个函数的导数。
5. Quotient Rule | 商法则
For a function written as a fraction where both numerator and denominator are functions of x, the quotient rule is used.
当一个函数写成分式形式，且分子和分母都是关于 x 的函数时，应使用商法则。
d/dx [u/v] = (u’v − u v’) / v²
If y = (x² + 1)/(x − 2), let u = x² + 1 and v = x − 2. Then u’ = 2x and v’ = 1, so dy/dx = [(2x)(x−2) − (x²+1)(1)] / (x−2)².
若 y = (x² + 1)/(x − 2)，设 u = x² + 1，v = x − 2。则 u’ = 2x，v’ = 1，因此 dy/dx = [(2x)(x−2) − (x²+1)(1)] / (x−2)²。
This simplifies to (2x² − 4x − x² − 1)/(x−2)² = (x² − 4x − 1)/(x−2)². Pay close attention to signs when subtracting.
此式可化简为 (2x² − 4x − x² − 1)/(x−2)² = (x² − 4x − 1)/(x−2)²。做减法时务必注意符号。
6. Differentiating Standard Functions | 常见函数的导数
Edexcel Pure Mathematics requires fluency with the derivatives of standard functions. Knowing these results by heart saves time in examinations.
Edexcel 纯数学要求熟练掌推常见函数的导数。熟记这些结果可以在考试中节省时间。
| Function f(x) | Derivative f'(x) |
|---|---|
| xⁿ | n xⁿ⁻¹ |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
| eˣ | eˣ |
| eᵏˣ | k eᵏˣ |
| ln x | 1/x |
| aˣ | aˣ ln a |
For composite forms such as sin(ax + b) or ln(ax + b), use the chain rule together with these standard results. For example, d/dx [sin(2x)] = 2 cos(2x).
对于 sin(ax + b) 或 ln(ax + b) 等复合形式，应将这些标准结果与链式法则结合使用。例如，d/dx [sin(2x)] = 2 cos(2x)。
7. Parametric Differentiation | 参数方程求导
When x and y are both given in terms of a third variable, usually t, the curve is parametric. The gradient is found by dividing dy/dt by dx/dt.
当 x 和 y 都由第三个变量（通常是 t）表示时，曲线为参数形式。其梯度等于 dy/dt 除以 dx/dt。
dy/dx = (dy/dt) / (dx/dt)
If x = t² + 1 and y = 2t³, then dx/dt = 2t and dy/dt = 6t². Hence dy/dx = 6t² / 2t = 3t.
若 x = t² + 1，y = 2t³，则 dx/dt = 2t，dy/dt = 6t²。因此 dy/dx = 6t² / 2t = 3t。
This method is especially useful for curves that are difficult to express in the form y = f(x). At a specific value of t, you can find the tangent slope directly.
这种方法对于难以表示为 y = f(x) 的曲线特别有用。在特定的 t 值下，可以直接求出切线斜率。
8. Implicit Differentiation | 隐函数求导
Implicit differentiation is used when y cannot easily be made the subject of the equation. Each term involving y must be differentiated, then multiplied by dy/dx.
当 y 难以写成显式函数时，使用隐函数求导。每个含 y 的项求导后都要乘以 dy/dx。
d/dx [y²] = 2y dy/dx
For the circle x² + y² = 25, differentiating both sides gives 2x + 2y dy/dx = 0. Rearranging yields dy/dx = −x/y.
对于圆 x² + y² = 25，两边求导得 2x + 2y dy/dx = 0。整理得到 dy/dx = −x/y。
Edexcel often combines implicit differentiation with finding tangents or stationary points. Remember that stationary points still require dy/dx = 0.
Edexcel 常将隐函数求导与求切线或驻点结合考查。记住驻点仍然要求 dy/dx = 0。
9. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative is the derivative of the first derivative. It describes how the gradient itself is changing and indicates the concavity of the graph.
二阶导数是导数的导数。它描述梯度本身的变化情况，并反映图像的凹凸性。
d²y/dx² = d/dx [dy/dx]
If d²y/dx² > 0 at a point, the curve is concave up. If d²y/dx² < 0, the curve is concave down.
如果某点处 d²y/dx² > 0，曲线向上凹；如果 d²y/dx² < 0，曲线向下凹。
For y = x³ − 3x, the first derivative is 3x² − 3 and the second derivative is 6x. At x = 1, d²y/dx² = 6 > 0, so the stationary point is a local minimum.
对于 y = x³ − 3x，一阶导数为 3x² − 3，二阶导数为 6x。在 x = 1 处，d²y/dx² = 6 > 0，因此该驻点为局部极小值。
10. Equations of Tangents and Normals | 切线与法线方程
A tangent is a straight line that touches a curve at a given point and has the same gradient as the curve at that point. Its gradient is found by evaluating dy/dx.
切线是与曲线在某一点相接触并且在该点斜率与曲线相同的直线。其斜率可通过计算 dy/dx 求得。
y − y₁ = m(x − x₁)
The normal is perpendicular to the tangent. If the tangent gradient is m, the normal gradient is −1/m.
法线垂直于切线。如果切线斜率为 m，则法线斜率为 −1/m。
For y = x² at the point (3, 9), dy/dx = 2x gives m = 6. The tangent is y − 9 = 6(x − 3), and the normal is y − 9 = −1/6(x − 3).
对于 y = x² 在点 (3, 9) 处，dy/dx = 2x 得到 m = 6。切线为 y − 9 = 6(x − 3)，法线为 y − 9 = −1/6(x − 3)。
11. Stationary Points and Optimisation | 驻点与优化问题
Stationary points occur where the first derivative is zero. They can be classified as maximum points, minimum points or points of inflection.
驻点出现在一阶导数为零的位置。它们可以分为极大值点、极小值点或拐点。
dy/dx = 0
To classify, use the second derivative test when possible: d²y/dx² < 0 gives a maximum, d²y/dx² > 0 gives a minimum, and d²y/dx² = 0 requires checking the sign change of dy/dx.
分类时尽量使用二阶导数判别法：d²y/dx² < 0 为极大值，d²y/dx² > 0 为极小值，若 d²y/dx² = 0 则需要检查 dy/dx 的符号变化。
Optimisation questions ask you to maximise or minimise a real-world quantity such as volume, area or cost. Write the quantity as a function of one variable, differentiate, solve for zero, and justify the nature of the point.
优化题要求最大化或最小化现实中的量，如体积、面积或成本。将量写成一个变量函数，求导，令其为零，并说明驻点的性质。
12. Rates of Change in Context | 相关变化率
Connected rates of change problems link the rate of change of one variable to another using the chain rule. This is a common modelling skill in Edexcel Pure Mathematics.
相关变化率问题利用链式法则将一个变量的变化率与另一个变量联系起来。这是 Edexcel 纯数学中常见的建模技能。
dV/dt = dV/dr × dr/dt
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