📚 Mastering Differentiation for Edexcel A-Level Pure Mathematics | 掌握爱德思 A-Level 纯数学微分
Differentiation is one of the most heavily examined topics in Edexcel A-Level Pure Mathematics, appearing in nearly every Paper 1 and Paper 2. A strong command of differentiation rules, combined with the ability to interpret gradients and rates of change, will help you secure high marks on both routine and applied questions.
微分是爱德思 A-Level 纯数学中考频最高的主题之一，几乎出现在每份 Paper 1 和 Paper 2 中。扎实掌握微分法则，并且能够解释梯度和变化率，将帮助你在常规题和应用题中稳拿高分。
1. The Derivative as a Limit | 导数作为极限
ive is formally introduced as the limit of the average rate of change. If y = f(x), then f' amined through first-principles questions, especially for simple powers such as f(x) = x².
在爱德思 A-Level 中，导数正式定义为平均变化率的极限。如果 y = f(x)，那么 f'(x) = lim(h→0) [f(x+h) - f(x)] / h。这一定义通常通过第一性原理题来考查，特别是针对简单幂函数，如 f(x) = x²。
(x+h)² = x² + 2xh + h², subtract f(x), divide by h, and then let h tend to 0. You obtain f'(x) = 2x. Showing each algebraic step clearly is essential for full marks.
例如，对 y = x² 从第一性原理求导，展开 (x+h)² = x² + 2xh + h²，减去 f(x)，除以 h，然后令 h 趋于 0，即可得到 f'(x) = 2x。清晰地展示每一步代数过程是获得满分的关键。
2. Basic Differentiation Rules | 基本微分法则
naxⁿ⁻¹. This rule works for any real constant n, including negative and fractional powers.
幂法则是最常用的微分工具。对于 y = axⁿ，先乘以指数 n，再将指数减一，得到 dy/dx = naxⁿ⁻¹。这一法则适用于任意实数指数 n，包括负数指数和分数指数。
- If y = 5x³, then dy/dx = 15x².
- 如果 y = 5x³，那么 dy/dx = 15x²。
- If y = 4x⁻², then dy/dx = -8x⁻³.
- 如果 y = 4x⁻²，那么 dy/dx = -8x⁻³。
- Rewrite y = √x as y = x^½ before differentiating, so dy/dx = ½x⁻^½.
- 先把 y = √x 改写为 y = x^½ 再求导，得到 dy/dx = ½x⁻^½。
The derivative of a constant is always zero, because a constant function has no rate of change. Also, the derivative of x itself is 1.
常数的导数始终为零，因为常数函数没有变化率。此外，x 本身的导数为 1。
3. Chain Rule for Composite Functions | 复合函数链式法则
y/dx = dy/du × du/dx, where u is an inner function chosen to simplify the differentiation.
许多爱德思考题要求对形如 y = [g(x)]ⁿ 或 y = sin(kx) 的函数使用链式法则。链式法则表明 dy/dx = dy/du × du/dx，其中 u 是为了简化求导而选择的内层函数。
If y = [f(x)]ⁿ, then dy/dx = n[f(x)]ⁿ⁻¹ f'(x)
如果 y = [f(x)]ⁿ，那么 dy/dx = n[f(x)]ⁿ⁻¹ f'(x)
x² + 5. Then y = u⁴, dy/du = 4u³, du/dx = 6x, so dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³.
例如，如果 y = (3x² + 5)⁴，设 u = 3x² + 5。那么 y = u⁴，dy/du = 4u³，du/dx = 6x，因此 dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³。
4. Product and Quotient Rules | 乘积法则与商法则
When two functions are multiplied or divided, the product and quotient rules must be used. These appear regularly in Edexcel Papers, often combined with trigonometric, exponential, or logarithmic functions.
当两个函数相乘或相除时，必须使用乘积法则和商法则。它们经常出现在爱德思试卷中，并且常常与三角函数、指数函数或对数函数结合考查。
Product rule: If y = uv, then dy/dx = u dv/dx + v du/dx
乘积法则：如果 y = uv，那么 dy/dx = u dv/dx + v du/dx
Quotient rule: If y = u/v, then dy/dx = (v du/dx - u dv/dx) / v²
商法则：如果 y = u/v，那么 dy/dx = (v du/dx - u dv/dx) / v²
For example, if y = x²eˣ, let u = x² and v = eˣ. Then dy/dx = x² × eˣ + eˣ × 2x = xeˣ Be careful to keep the order in the quotient rule, because subtraction is not commutative.
例如，如果 y = x²eˣ，设 u = x²，v = eˣ。那么 dy/dx = x² × eˣ + eˣ × 2x = xeˣ(x + 2)。商法则中要注意保持顺序，因为减法不满足交换律。
5. Differentiating Exponential and Logarithmic Functions | 指数与对数函数微分
The exponential function eˣ has the unique property that its derivative is itself. More generally, if y = ekx, then dy/dx = kekx. This appears constantly in growth and decay contexts.
指数函数 eˣ 具有导数等于自身的独特性质。更一般地，如果 y = ekx，那么 dy/dx = kekx。这在增长和衰减问题中频繁出现。
as well, since ln(kx) = ln k + ln x and ln k is a constant term that differentiates to 0.
对于自然对数，ln x 的导数是 1/x。如果 y = ln(kx)，那么 dy/dx 也等于 1/x，因为 ln(kx) = ln k + ln x，而 ln k 是常数项，求导后为 0。
d/dx (eˣ) = eˣ, d/dx (ekx) = kekx, d/dx (ln x) = 1/x
d/dx (eˣ) = eˣ，d/dx (ekx) = kekx，d/dx (ln x) = 1/x
6. Implicit Differentiation | 隐函数微分
ate both sides with respect to x, treating y as a function of x and using the chain rule on any y term.
隐函数微分用于 y 没有显式表示为 x 的函数的情况。例如，方程 x² + y² = 25 定义了一个圆。对两边关于 x 求导，将 y 视为 x 的函数，并对所有含 y 的项使用链式法则。
. You must then substitute the given coordinates to find the gradient at a specific point.
对于 x² + y² = 25，求导得到 2x + 2y dy/dx = 0，因此 dy/dx = -x/y。接着必须代入给定坐标来求某一点处的梯度。
Implicit differentiation questions often require collecting all dy/dx terms on one side and factorising. It is also common to combine this skill with finding tangents and normals to curves defined implicitly.
隐函数微分题通常要求把所有的 dy/dx 项移到一边并进行因式分解。这种技巧也常与求隐式定义的曲线的切线和法线相结合。
7. Parametric Differentiation | 参数方程微分
ent is found by dividing the derivative of y with respect to t by the derivative of x with respect to t.
当曲线由参数方程 x = f(t) 和 y = g(t) 给出时，梯度通过将 y 对 t 的导数除以 x 对 t 的导数来求得。
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (dy/dt) / (dx/dt)
vative requires a further step: differentiate dy/dx with respect to t and divide by dx/dt.
例如，如果 x = t² 且 y = 2t + 1，那么 dx/dt = 2t，dy/dt = 2，因此 dy/dx = 2 / 2t = 1/t。二阶导数需要进一步运算：对 dy/dx 关于 t 求导，再除以 dx/dt。
d²y/dx² = d/dt (dy/dx) / (dx/dt)
d²y/dx² = d/dt (dy/dx) / (dx/dt)
8. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative, written d²y/dx² or f''(x), measures the rate of change of the gradient. It is used to classify stationary points and to determine whether a curve is concave up or concave down on an interval.
二阶导数，写作 d²y/dx² 或 f''(x)，用于衡量梯度的变化率。它被用来判断驻点的性质，以及曲线在某一区间是向上凹还是向下凹。
ximum. If d²y/dx² = 0, the test is inconclusive, and you should use a gradient sign table.
在驻点处，即 dy/dx = 0 时，计算 d²y/dx²。如果 d²y/dx² > 0，该点为局部极小值；如果 d²y/dx² < 0，该点为局部极大值；如果 d²y/dx² = 0，则无法判断，应使用梯度符号表进行检验。
9. Tangents, Normals and Stationary Points | 切线、法线与驻点
The gradient of a tangent at a point equals dy/dx evaluated at that point. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient.
切线在某一点的斜率等于 dy/dx 在该点的值。法线垂直于切线，因此法线的斜率是切线斜率的负倒数。
m_tangent = dy/dx, mnormal = -1 / m_tangent
m_切线 = dy/dx，m_法线 = -1 / m_切线
, substitute back into the original equation to find the corresponding y-coordinates. Then classify each point as a maximum, minimum, or point of inflection.
驻点出现在 dy/dx = 0 的位置。求出 x 坐标后，代回原方程以求得对应的 y 坐标。然后将每个点分类为极大值点、极小值点或拐点。
10. Optimisation Problems | 优化问题
ive function, differentiate, set dy/dx = 0, and verify the nature of the stationary point.
优化题要求你求某个量的最大值或最小值，例如面积、体积、成本或距离。关键步骤是用单一变量建立模型，构造目标函数，求导，令 dy/dx = 0，并验证驻点性质。
Always state the practical domain of the variable. For example, if x represents a length, then x must be positive, and the endpoints of the feasible interval should be checked if the stationary point is outside the allowed range.
始终说明变量的实际取值范围。例如，如果 x 表示长度，那么 x 必须为正数，而且如果驻点不在允许区间内，还需检查可行区间的端点。
Once the maximum or minimum is found, interpret the answer in context and include the correct units. Edexcel mark schemes award method marks for differentiation, solving, and justification, even if the final numerical answer is wrong.
一旦求出最大值或最小值，应结合实际问题解释答案并写明正确单位。即使最终数值答案有误，爱德思评分方案也会对求导、求解和验证过程给予方法分。
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