📚 Mastering Differentiation for Edexcel A-Level Maths | Edexcel A-Level 数学：掌握微分法
Differentiation is one of the most heavily examined topics in the Edexcel A-Level Mathematics specification. It appears in Pure Mathematics 1 and 2, and it underpins applications such as tangents, normals, stationary points, optimisation and connected rates of change.
微分是 Edexcel A-Level 数学大纲中考查最频繁的主题之一。它出现在 Pure Mathematics 1 和 2 中，并且是切线、法线、驻点、最优化以及相关变化率等应用的基础。
1. The Gradient Function and First Principles | 梯度函数与第一原理
The derivative f' t can be defined as a limit of the gradient of a chord as the chord gets infinitely short.
导数 f'(x) 测量曲线 y = f(x) 在任意给定点的梯度。它可定义为当割线无限变短时，割线梯度的极限。
f'(x) = lim (h → 0) [f(x + h) − f(x)] / h
This formula is called differentiation from first principles. In the exam you may be asked to use it for simple polynomials such as x² or x³.
这个公式称为从第一原理求导。考试中可能要求你对简单多项式（如 x² 或 x³）使用该公式。
2. The Power Rule and Basic Derivatives | 幂法则与基本导数
then dy/dx = n xⁿ⁻¹. To apply it, multiply by the power, then subtract one from the power.
对于任意实数次幂 n，幂法则指出：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。应用时先乘以幂指数，再将指数减一。
Common results are:
常见结果如下：
- y = x⁵ ⇒ dy/dx = 5x⁴
- y = 3x² ⇒ dy/dx = 6x
- y = 7 ⇒ dy/dx = 0
- y = 4x ⇒ dy/dx = 4
Remember that constants can be taken outside the derivative, and the derivative of a constant alone is zero.
记住，常数可以提到导数符号外，单独常数的导数为零。
3. From First Principles at a Point | 在一点处从第一原理求导
You can also find the gradient at a specific point x = a by evaluating f' definition. It often requires expanding brackets and cancelling the h in the denominator.
你还可以使用极限定义计算特定点 x = a 处的梯度 f'(a)。这通常需要展开括号并约去分母中的 h。
For f(x) = x², f' lim (h → 0) [(x + h)² − x²] / h = lim (h → 0) [2xh + h²] / h = lim (h → 0) (2x + h) = 2x.
对于 f(x) = x²，f' = lim (h → 0) [(x + h)² − x²] / h = lim (h → 0) [2xh + h²] / h = lim (h → 0) (2x + h) = 2x。
4. The Chain Rule | 链式法则
when differentiating a function of a function. If y = [u(x)]ⁿ then dy/dx = n[u(x)]ⁿ⁻¹ × u'(x).
链式法则用于复合函数求导。若 y = [u(x)]ⁿ，则 dy/dx = n[u(x)]ⁿ⁻¹ × u'(x)。
A simple way to remember it is: differentiate the outer function, then multiply by the derivative of the inner function.
一个简单的记忆方法是：先对外层函数求导，再乘以内层函数的导数。
dy/dx = (dy/du) × (du/dx)
Example: y = (3x² + 1)⁵ gives dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.
示例：y = (3x² + 1)⁵，则 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。
5. The Product and Quotient Rules | 乘积法则与商法则
For two functions u(x) and v(x), the product rule states:
对于两个函数 u(x) 和 v(x)，乘积法则为：
y = uv ⇒ dy/dx = u'v + uv'
The quotient rule states:
商法则为：
y = u/v ⇒ dy/dx = (u'v − uv') / v²
When using the quotient rule, keep the denominator squared and be careful with the negative sign in the numerator.
使用商法则时，分母要变成原分母的平方，并注意分子中的负号。
Example: y = x² sin x gives dy/dx = 2x sin x + x² cos x.
示例：y = x² sin x，则 dy/dx = 2x sin x + x² cos x。
6. Tangents and Normals | 切线与法线
The gradient of the tangent to y = f(x) at x = a is m = f'(a). The equation of the tangent is y − f(a) = m(x − a).
曲线 y = f(x) 在 x = a 处切线的梯度为 m = f'(a)。切线方程为 y − f(a) = m(x − a)。
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal −1/m. Use the same point (a, f(a)) to form the normal equation.
法线垂直于切线，因此其梯度为负倒数 −1/m。使用同一点 (a, f(a)) 建立法线方程。
Example: For y = x² at x = 1, f'(1) = 2. Tangent: y − 1 = 2(x − 1). Normal: y − 1 = −1/2 (x − 1).
示例：对于 y = x² 在 x = 1 处，f'(1) = 2。切线：y − 1 = 2(x − 1)。法线：y − 1 = −1/2 (x − 1)。
7. Stationary Points and the Second Derivative Test | 驻点与二阶导数判别法
0. To classify them, you can use the second derivative d²y/dx² or a sign table for dy/dx.
驻点出现在 dy/dx = 0 的位置。判断驻点类型，可以使用二阶导数 d²y/dx²，也可以使用 dy/dx 的符号表。
- d²y/dx² > 0 ⇒ local minimum.
- d²y/dx² < 0 ⇒ local maximum.
- d²y/dx² = 0 ⇒ test is inconclusive; use a sign table.
If the second derivative is positive, the curve is convex and the point is a minimum; if negative, the curve is concave and the point is a maximum.
如果二阶导数为正，曲线是凸的，该点为极小值；如果为负，曲线是凹的，该点为极大值。
8. Increasing and Decreasing Functions | 增函数与减函数
A function is increasing on an interval if f'(x) > 0 for all x in that interval. It is decreasing if f'(x) < 0.
若在某个区间内 f'(x) > 0，则函数在该区间上递增；若 f'(x) < 0，则函数递减。
To find intervals of increase or decrease, solve f'(x) > 0 or f'(x) < 0. Exam questions often combine this with quadratic inequalities.
要求递增或递减区间，可分别解不等式 f'(x) > 0 或 f'(x) < 0。考试题常将这一内容与二次不等式结合。
Example: f(x) = x³ − 3x gives f' − 3 = 3(x² − 1). Increasing when x < −1 or x > 1; decreasing when −1 < x < 1.
示例：f(x) = x³ − 3x，f'(x) = 3x² − 3 = 3(x² − 1)。当 x < −1 或 x > 1 时递增；当 −1 < x < 1 时递减。
9. Optimisation Problems | 最优化问题
Optimisation problems ask you to maximise or minimise a quantity such as area, volume or cost. You first write the quantity as a function of one variable, then set its derivative equal to zero.
最优化问题要求你最大化或最小化面积、体积或成本等量。首先将该量写成单变量函数，然后令其导数等于零。
Always verify that your stationary point is the required maximum or minimum, using the second derivative test or a sign test.
始终要验证所得驻点是否为所需的最大值或最小值，可使用二阶导数判别法或符号判别法。
linking them; eliminate one variable; differentiate; solve dy/dx = 0; justify the nature.
典型步骤为：定义变量；建立变量间关系；消去一个变量；求导；解 dy/dx = 0；说明驻点性质。
10. Connected Rates of Change | 相关变化率
time t, their rates of change are connected by the chain rule: dy/dt = (dy/dx) × (dx/dt).
如果两个量 x 和 y 都随时间 t 变化，它们的变化率由链式法则联系：dy/dt = (dy/dx) × (dx/dt)。
In a typical problem, you are given one rate and asked to find another. Identify the missing derivative dy/dx, compute it, then multiply by the known rate dx/dt.
在典型问题中，给出一个变化率，要求另一个。找出缺少的导数 dy/dx，计算它，再乘以已知变化率 dx/dt。
pherical balloon expands so dr/dt = 0.5 cm s⁻¹. Since V = 4/3 π r³, dV/dt = 4π r² × dr/dt.
示例：一个球形气球膨胀，dr/dt = 0.5 cm s⁻¹。由于 V = 4/3 π r³，dV/dt = 4π r² × dr/dt。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图