📚 Edexcel A-Level Pure Mathematics: Differentiation Rules and Applications | 爱德思A-Level纯数学：微分法则与应用
Differentiation is one of the most heavily examined topics in Edexcel A-Level Mathematics. It connects algebraic techniques with real-world rates of change, optimisation and curve analysis, and it appears in almost every paper of Pure Mathematics 1 and Pure Mathematics 2.
微分是爱德思A-Level数学中考查最频繁的主题之一。它把代数技巧与现实中的变化率、优化问题和曲线分析联系在一起，几乎出现在纯数学1和纯数学2的每一份试卷中。
This revision guide covers first principles, the main differentiation rules, implicit and parametric methods, connected rates and stationary points. Each section is written as an exam-focused summary of the techniques Edexcel candidates need to apply confidently, with worked reasoning and the errors that most often cost marks.
本复习指南涵盖第一性原理、主要求导法则、隐函数与参数方法、相关变化率和驻点。每一节都以考试重点为导向，总结爱德思考生需要熟练运用的技巧，并说明最容易失分的错误。
1. Gradient and Derivative from First Principles | 梯度与第一性原理求导
The derivative f'(x) measures the instantaneous rate of change of a function at a point. From first principles, it is the limit of average gradients of chords as the interval tends to zero. A chord between (x, f(x)) and (x + h, f(x + h)) has gradient [f(x + h) - f(x)] / h, and letting h shrink to zero turns this average gradient into the gradient of the tangent.
导数 f'(x) 度量函数在某一点的瞬时变化率。从第一性原理出发，它是当区间趋于零时弦的平均梯度的极限。连接 (x, f(x)) 与 (x + h, f(x + h)) 的弦，其梯度为 [f(x + h) - f(x)] / h；令 h 趋近于零，这个平均梯度就变成切线的梯度。
f'(x) = lim (h → 0) [ f(x + h) - f(x) ] / h
For example, differentiating f(x) = x² from first principles gives f' ch as f(x) = 1 / x, where the algebra requires a common denominator before the limit is taken.
例如，对 f(x) = x² 从第一性原理求导可得 f'(x) = 2x，因为差商可化简为 2x + h，然后令 h 趋于零。相同的极限论证也适用于 x³，以及 f(x) = 1 / x 这类简单情形，只是取极限前需要先通分。
In Edexcel questions, first-principles proofs are usually required only for simple polynomial terms, but the limit interpretation underpins every differentiation rule you will use. A typical exam instruction is 'differentiate from first principles', which means you must show the difference quotient and the limiting step explicitly, not simply quote the power rule.
在爱德思考试题中，通常只要求证明简单多项式项的第一性原理，但极限意义是你将使用的每一条微分法则的基础。典型考题指令是 'differentiate from first principles'（用第一性原理求导），这意味着你必须明确写出差商和取极限的步骤，而不能直接引用幂法则。
2. Standard Derivatives and the Power Rule | 常见导数与幂法则
The power rule states that if f(x) = xⁿ for any real constant n, then f' ngle rule handles polynomials, roots and reciprocals once they are written as powers of x.
幂法则指出，如果 f(x) = xⁿ，其中 n 为任意实常数，则 f'(x) = n xⁿ⁻¹。该法则可线性推广：常数倍可直接提出，和可逐项求导。只要把根式和倒数写成 x 的幂，这一条法则就能处理多项式、根式和倒数。
d/dx (xⁿ) = n xⁿ⁻¹
careful with negative and fractional powers. For instance, d/dx (x⁻²) = -2x⁻³, and d/dx (√x) = d/dx (x^½) = ½ x−½ = 1 / (2√x). Rewriting every expression in index form before differentiating is the safest habit, especially when the question contains a fraction such as 3 / x⁴, which becomes 3x⁻⁴.
必须小心负指数和分数指数。例如 d/dx (x⁻²) = -2x⁻³；又如 d/dx (√x) = d/dx (x^½) = ½ x−½ = 1 / (2√x)。最稳妥的习惯是先化成指数形式再求导，尤其当题目中出现像 3 / x⁴ 这样的分式时，应改写为 3x⁻⁴。
- d/dx (constant) = 0, because a horizontal line has zero gradient.
- d/dx (x) = 1, the simplest linear case.
- d/dx (a xⁿ) = a n xⁿ⁻¹, where a is a constant.
- d/dx (xⁿ + xᵐ) = n xⁿ⁻¹ + m xᵐ⁻¹, differentiating term by term.
- d/dx (常数) = 0，因为水平线的梯度为零。
- d/dx (x) = 1，最简单的线性情形。
- d/dx (a xⁿ) = a n xⁿ⁻¹，其中 a 为常数。
- d/dx (xⁿ + xᵐ) = n xⁿ⁻¹ + m xᵐ⁻¹，逐项求导。
A common exam error is to subtract one from the coefficient instead of the index, or to forget that the derivative of a constant vanishes. Another is to leave a negative index in the final answer when the question asks for a form with positive powers.
常见失分点是把系数减一而不是把指数减一，或者忘记常数项导数为零。另一种错误是当题目要求用正指数形式作答时，最终答案仍保留负指数。
3. Standard Derivatives of Trigonometric and Exponential Functions | 三角与指数函数的常见导数
Edexcel expects you to know the derivatives of sine, cosine, tangent, eˣ and ln x, and to apply them inside composite functions. The results in the table below are given in the formula booklet, but you must know how to use them accurately under time pressure.
爱德思要求你掌握 sin、cos、tan、eˣ 和 ln x 的导数，并能在复合函数中熟练运用。下表结果虽在公式册中给出，但你必须在考试时间压力下准确使用它们。
| Function f(x) | Derivative f'(x) | Key note |
| sin x | cos x | Angles must be in radians. |
| cos x | -sin x | The minus sign is essential. |
| tan x | sec² x = 1 / cos² x | Proof uses the quotient rule. |
| eˣ | eˣ | The gradient equals the value. |
| ln x | 1 / x | Valid for x > 0 only. |
For example, d/dx (3 sin x - 2 eˣ) = 3 cos x - 2 eˣ. When these functions appear with a linear argument, such as sin(4x) or e5x produces an extra factor of the inner derivative, so d/dx (sin 4x) = 4 cos 4x and d/dx (e5x) = 5 e5x.
例如 d/dx (3 sin x - 2 eˣ) = 3 cos x - 2 eˣ。当这些函数带线性内层，如 sin(4x) 或 e5x 时，链式法则会带来内层导数的额外因子，因此 d/dx (sin 4x) = 4 cos 4x，d/dx (e5x) = 5 e5x。
4. The Chain Rule for Composite Functions | 复合函数的链式法则
The chain rule differentiates a function of a function. If y is a function of u and u is a function of x, then the rates multiply. In function notation, the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x).
链式法则用于对函数的函数求导。如果 y 是 u 的函数，而 u 是 x 的函数，那么变化率相乘。用函数记号表示，f(g(x)) 的导数是 f'(g(x)) 乘以 g'(x)。
dy/dx = (dy/du) × (du/dx)
du = 5u⁴ and du/dx = 6x, giving dy/dx = 30x(3x² + 1)⁴. The same structure applies to y = e2x + 1, where dy/dx = 2 e2x + 1, and to y = ln(5x + 2), where dy/dx = 5 / (5x + 2).
考虑 y = (3x² + 1)⁵。令 u = 3x² + 1，则 y = u⁵。于是 dy/du = 5u⁴，du/dx = 6x，因此 dy/dx = 30x(3x² + 1)⁴。同样的结构适用于 y = e2x + 1，其 dy/dx = 2 e2x + 1；也适用于 y = ln(5x + 2)，其 dy/dx = 5 / (5x + 2)。
- Identify the inner function u before differentiating.
- Differentiate the outer function with respect to u, keeping u intact.
- Multiply by du/dx, then substitute u back in terms of x.
- Simplify fully, especially when negative or fractional powers are involved.
- 先确定内层函数 u，再求导。
- 对外层函数关于 u 求导，保持 u 不变。
- 乘以 du/dx，再把 u 换回 x 的表达式。
- 彻底化简，尤其是涉及负指数或分数指数时。
The most frequent mistake is to differentiate the outer function and forget the inner derivative altogether. A second error is to substitute u back too early, which makes the algebra harder to track.
最常见的错误是只对外层函数求导而完全忘记内层导数。第二个错误是过早把 u 换回 x，使代数过程更难追踪。
5. The Product Rule | 乘积法则
lied together. If y = u v, where u and v are both functions of x, then the derivative is u'v + uv' le is not simply the product of the two derivatives, which is a very common misconception.
当两个可导函数相乘时使用乘积法则。若 y = u v，其中 u 和 v 都是 x 的函数，则导数为 u'v + uv'。该法则并不是两个导数直接相乘，这是非常常见的误解。
d/dx (u v) = u (dv/dx) + v (du/dx)
For y = x² eˣ, take u = x² and v = eˣ. Then u' = 2x and v' = eˣ, so dy/dx = 2x eˣ + x² eˣ = x eˣ it makes the location of stationary points easier to find in later parts of the question.
对于 y = x² eˣ，取 u = x²，v = eˣ，则 u' = 2x，v' = eˣ，所以 dy/dx = 2x eˣ + x² eˣ = x eˣ (x + 2)。通常需要把答案因式分解，因为这样在后续小题中更容易找到驻点位置。
roduct rule must be combined with the chain rule, giving dy/dx = 2x sin(3x) + 3x² cos(3x).
另一个标准例子是 y = x sin x，其导数为 dy/dx = sin x + x cos x。当某个因子本身是复合函数时，例如 y = x² sin(3x)，就必须把乘积法则与链式法则结合，得到 dy/dx = 2x sin(3x) + 3x² cos(3x)。
6. The Quotient Rule | 商法则
numerator and denominator are both functions of x. If y = u / v, then the derivative is (u'v - uv' v². The order in the numerator matters: reversing it changes the sign of the whole answer.
商法则用于对分子和分母都是 x 的函数的分数求导。若 y = u / v，则导数为 (u'v - uv') / v²。分子中的顺序很重要：颠倒顺序会使整个答案的符号改变。
d/dx (u / v) = (v (du/dx) - u (dv/dx)) / v²
cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x, using the identity sin² x + cos² x = 1.
经典应用是证明 tan x 的导数。把 tan x 写成 sin x / cos x，令 u = sin x，v = cos x，则 x · cos x - sin x · (-sin x)) / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x，其中用到恒等式 sin² x + cos² x = 1。
w the x terms cancel, leaving a negative constant numerator, which tells you the function is always decreasing on each side of its asymptote.
对于多项式商，如 y = (2x + 1) / (x - 3)，法则给出 dy/dx = [2(x - 3) - (2x + 1)(1)] / (x - 3)² = (2x - 6 - 2x - 1) / (x - 3)² = -7 / (x - 3)²。注意 x 项相消后留下负常数分子，这说明该函数在渐近线两侧都是单调递减的。
7. Tangents, Normals and Rates of Change | 切线、法线与变化率
gives the gradient of the tangent to a curve at a point. Once you have the gradient m = f' normal is perpendicular to the tangent, so its gradient is -1 / m, provided m is not zero.
导数给出曲线在某点处切线的梯度。求出梯度 m = f'(x₁) 和点 (x₁, y₁) 后，切线方程为 y - y₁ = m(x - x₁)。法线与切线垂直，因此其梯度为 -1 / m，前提是 m 不为零。
mnormal × mtangent = -1
The derivative also models rates of change in applied contexts. If s(t) is displacement, then ds/dt is velocity and d²s/dt² is acceleration. If V(t) is the volume of a expanding balloon, then dV/dt is the rate of increase of volume with respect to time, measured in cm³ s⁻¹.
导数也用于建模应用情境中的变化率。若 s(t) 是位移，则 ds/dt 是速度，d²s/dt² 是加速度。若 V(t) 是膨胀气球的体积，则 dV/dt 是体积关于时间的变化率，单位为 cm³ s⁻¹。
- Differentiate, then substitute the x-coordinate to obtain the gradient.
- Use the original equation to find the y-coordinate of the point.
- For the normal, take the negative reciprocal of the tangent gradient.
- Always state units in rate-of-change questions.
- 先求导，再代入横坐标得到梯度。
- 用原方程求出该点的纵坐标。
- 求法线时，取切线梯度的负倒数。
- 在变化率题目中务必写出单位。
A frequent error is to use the x-coordinate in the derivative but forget to find the corresponding y-coordinate, or to use the gradient of the tangent in the normal equation. Both lead to a correct method with an incorrect final answer, which usually loses the accuracy mark.
常见错误是只把横坐标代入导数，却忘记求对应的纵坐标；或者在法线方程中误用切线梯度。两者都会导致方法正确但最终答案错误，通常失掉精度分。
8. Stationary Points, Maxima and Minima | 驻点、极大与极小
A stationary point occurs where f' meaning the tangent is horizontal. To classify the point, evaluate the second derivative f''(x): if f''(x) > 0 the point is a local minimum, if f''(x) < 0 it is a local maximum, and if f''(x) = 0 the test is inconclusive and you must examine the sign of f'(x) on either side.
驻点出现在 f'(x) = 0 处，即切线水平。判断类型时计算二阶导数 f''(x)：若 f''(x) > 0，该点为局部极小；若 f''(x) < 0，该点为局部极大；若 f''(x) = 0，则此判别法失效，必须检查 f'(x) 在两侧的符号。
f'(x) = 0 and f''(x) > 0 → local minimum
ive at x = 1 and negative at x = -1. Therefore (1, -2) is a local minimum and (-1, 2) is a local maximum.
考虑 y = x³ - 3x。则 dy/dx = 3x² - 3 = 3(x - 1)(x + 1)，所以驻点出现在 x = 1 和 x = -1。二阶导数为 d²y/dx² = 6x，在 x = 1 处为正，在 x = -1 处为负。因此 (1, -2) 是局部极小点，(-1, 2) 是局部极大点。
In optimisation problems, you must also consider the context. A length cannot be negative, and a maximum may occur at an endpoint of the valid interval rather than at a stationary point. Always re-read the question to confirm the range of values the variable can take.
在优化问题中，还必须考虑实际背景。长度不能为负，最大值也可能出现在有效区间的端点而非驻点。务必重新阅读题目，确认变量可取的范围。
9. Implicit Differentiation | 隐函数求导
every term with respect to x, remembering that differentiating a term in y produces dy/dx by the chain rule.
当 y 没有明确写成 x 的函数，而是由 x² + y² = 25 这样的关系式定义时，就要用隐函数求导。对每一项关于 x 求导，并记住：对含 y 的项求导会因链式法则产生 dy/dx。
d/dx (y²) = 2y (dy/dx)
rresponds to vertical tangents. The gradient of the tangent at (3, 4) is therefore -3 / 4.
对 x² + y² = 25 求导得 2x + 2y dy/dx = 0，所以 dy/dx = -x / y。该结果在圆上除 y = 0（对应竖直切线）外的每一点都成立。因此点 (3, 4) 处切线梯度为 -3 / 4。
Products and quotients involving both x and y require extra care. For example, differentiating x y with respect to x gives y + x dy/dx, using the product rule, and differentiating sin y gives cos y dy/dx. After differentiating, collect all dy/dx terms on one side and factorise to isolate the derivative.
同时含 x 和 y 的乘积与商需要格外小心。例如，对 x y 关于 x 求导得 y + x dy/dx（用乘积法则）；对 sin y 求导得 cos y dy/dx。求导后，把所有含 dy/dx 的项移到一边并因式分解，从而解出导数。
10. Parametric Differentiation | 参数方程求导
is found by dividing the two derivatives with respect to the parameter. This avoids eliminating t, which is often impossible or messy.
当曲线由 x = x(t) 和 y = y(t) 参数式定义时，梯度可通过两个关于参数的导数相除得到。这样可以避免消去 t，而消去 t 往往不可能或非常繁琐。
dy/dx = (dy/dt) ÷ (dx/dt)
1, so dy/dx = (3t² - 1) / (2t). The tangent is horizontal when dy/dt = 0, giving t = ±1 / √3, and vertical when dx/dt = 0, giving t = 0.
对于曲线 x = t²，y = t³ - t，有 dx/dt = 2t，dy/dt = 3t² - 1，所以 dy/dx = (3t² - 1) / (2t)。当 dy/dt = 0 时切线水平，得 t = ±1 / √3；当 dx/dt = 0 时切线竖直，得 t = 0。
(dx/dt). This is a standard Edexcel extension result. A common mistake is to differentiate dy/dx with respect to x directly; the correct method differentiates with respect to t and then divides by dx/dt.
参数曲线的二阶导数为 d²y/dx² = [d/dt (dy/dx)] / (dx/dt)。这是爱德思考试中的标准拓展结论。常见错误是直接对 dy/dx 关于 x 求导；正确方法是先关于 t 求导，再除以 dx/dt。
11. Connected Rates of Change | 相关变化率
Connected rates of change problems link two derivatives through the chain rule. If a quantity y depends on x, and x changes with time t, then the rate of change of y is the product of two familiar derivatives.
相关变化率问题通过链式法则把两个导数联系起来。若量 y 依赖于 x，而 x 随时间 t 变化，则 y 的变化率等于两个已知导数的乘积。
dy/dt = (dy/dx) × (dx/dt)
2π r dr/dt. If dr/dt = 0.5 cm s⁻¹ and r = 10 cm, then dA/dt = 2π × 10 × 0.5 = 10π cm² s⁻¹.
标准例子是半径匀速增大的圆形水波。面积为 A = π r²，所以 dA/dr = 2π r，dA/dt = 2π r dr/dt。若 dr/dt = 0.5 cm s⁻¹ 且 r = 10 cm，则 dA/dt = 2π × 10 × 0.5 = 10π cm² s⁻¹。
