Calculus Essentials: Differentiation and Integration | 微积分核心：微分与积分

📚 Calculus Essentials: Differentiation and Integration | 微积分核心：微分与积分

Edexcel A Level Mathematics rewards students who can move flexibly between algebraic, graphical and numerical representations. Calculus brings together limits, gradients, area and motion, so a strong command of differentiation and integration is essential for both Pure Mathematics papers. This revision guide covers the key techniques, standard results and common applications that examiners expect.

Edexcel A Level 数学要求学生在代数、图形和数值表示之间灵活转换。微积分将极限、斜率、面积与运动联系起来，因此熟练掌握微分与积分对两张纯数学试卷都至关重要。本复习指南涵盖关键技巧、标准结果以及考官重点考查的常见应用。

1. First Principles and the Gradient Function | 第一原理与梯度函数

The derivative of a function f(x) is defined from first principles as the limit of the gradient of a chord as the second point approaches the first. This definition underpins every differentiation rule, so Edexcel exam questions sometimes ask you to prove a derivative from this limit rather than using the standard rule.

函数 f(x) 的导数从第一原理定义为当第二个点趋近第一个点时弦的斜率的极限。这一定义支撑着所有微分法则，因此 Edexcel 考试有时会要求你从该极限证明导数，而不是直接使用标准公式。

f'(x) = limₕ→₀ [ f(x+h) − f(x) ] ÷ h

For example, if f(x) = x², then f(x+h) = x² + 2xh + h². The difference quotient becomes (2xh + h²) ÷ h = 2x + h, so as h → 0 the derivative is f'(x) = 2x. You should be able to repeat this argument for simple powers such as x³ and for constant multiples.

例如，若 f(x) = x²，则 f(x+h) = x² + 2xh + h²。差分商变为 (2xh + h²) ÷ h = 2x + h，因此当 h → 0 时，导数为 f'(x) = 2x。你应当能够对 x³ 以及常数倍等简单幂函数重复这一推导过程。

2. Basic Differentiation Rules | 基本微分法则

The power rule is the most frequently used tool in A Level differentiation. It states that to differentiate xⁿ, multiply by the power and reduce the power by one. Constants differentiate to zero, and a sum differentiates term by term, so you can handle polynomials very quickly.

幂函数法则是 A Level 微分中最常用的工具。它表示求 xⁿ 的导数时，先乘以指数，再将指数减一。常数求导为零，和式逐项求导，因此你可以非常快速地处理多项式。

f(x)	f'(x)
x⁵	5x⁴
3x²	6x
7	0
2x³ − 4x + 9	6x² − 4

The general rule d/dx (xⁿ) = nxⁿ⁻¹ also works for negative and fractional powers after rewriting terms such as 1/x² as x⁻² and √x as x¹/². This is especially important when differentiating rational or surd functions.

一般公式 d/dx (xⁿ) = nxⁿ⁻¹ 也适用于负指数和分数指数，只需将 1/x² 改写为 x⁻²，将 √x 改写为 x¹/²。这在求有理函数或根式函数的导数时尤为重要。

3. Chain, Product and Quotient Rules | 链式法则、乘法法则与除法法则

Composite functions require the chain rule. If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. In practice, differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function.

复合函数需要使用链式法则。如果 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。实际操作中，先对外层函数求导并保持内层函数不变，再乘以内层函数的导数。

y = (3x² + 1)⁵ ⇒ dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴

The product rule is used for two functions multiplied together: if y = uv, then dy/dx = u’v + uv’. The quotient rule applies to a function divided by another: if y = u/v, then dy/dx = (vu’ − uv’) ÷ v². You must know both formulas and recognise when to use them.

乘法法则用于两个函数相乘：若 y = uv，则 dy/dx = u’v + uv’。除法法则用于一个函数除以另一个函数：若 y = u/v，则 dy/dx = (vu’ − uv’) ÷ v²。你必须记住这两个公式并能识别何时使用它们。

For y = x² sin x, the product rule gives dy/dx = 2x sin x + x² cos x. For y = x ÷ (x + 1), the quotient rule gives [(x + 1)(1) − x(1)] ÷ (x + 1)² = 1 ÷ (x + 1)².

对于 y = x² sin x，乘法法则给出 dy/dx = 2x sin x + x² cos x。对于 y = x ÷ (x + 1)，除法法则给出 [(x + 1)(1) − x(1)] ÷ (x + 1)² = 1 ÷ (x + 1)²。

4. Higher Derivatives and Rates of Change | 高阶导数与变化率

The second derivative d²y/dx² is the derivative of dy/dx and measures how the gradient itself is changing. In kinematics, if displacement s is expressed as a function of time t, then velocity v = ds/dt and acceleration a = dv/dt = d²s/dt².

二阶导数 d²y/dx² 是 dy/dx 的导数，它衡量斜率本身的变化情况。在运动学中，如果位移 s 表示为时间 t 的函数，那么速度 v = ds/dt，加速度 a = dv/dt = d²s/dt²。

For example, if s(t) = t³ − 3t² + 2, then v = s'(t) = 3t² − 6t and a = v'(t) = 6t − 6. Substituting a specific time into these expressions gives the instantaneous velocity and acceleration, and setting a = 0 tells you when the velocity is stationary.

例如，若 s(t) = t³ − 3t² + 2，则 v = s'(t) = 3t² − 6t，a = v'(t) = 6t − 6。将具体时间代入这些表达式可得到瞬时速度和瞬时加速度，令 a = 0 可判断速度何时达到驻定状态。

5. Tangents, Normals and Optimisation | 切线、法线与最优化

The value dy/dx at a point is the gradient of the tangent to the curve at that point. The tangent line has equation y − y₁ = m(x − x₁), where m = dy/dx evaluated at x = x₁. The normal is perpendicular to the tangent, so its gradient is −1/m.

函数在某点的 dy/dx 值就是该点处切线的斜率。切线方程为 y − y₁ = m(x − x₁)，其中 m 是在 x = x₁ 处求得的 dy/dx。法线与切线垂直，因此其斜率为 −1/m。

For y = x² at x = 3, the gradient is m = 6 and y₁ = 9, so the tangent is y − 9 = 6(x − 3). The normal has gradient −1/6 and equation y − 9 = −(1/6)(x − 3). This type of question is frequently linked with coordinate geometry and circles.

对于 y = x² 在 x = 3 处，斜率为 m = 6，y₁ = 9，因此切线为 y − 9 = 6(x − 3)。法线斜率为 −1/6，方程为 y − 9 = −(1/6)(x − 3)。这类问题经常与坐标几何和圆的内容结合考查。

6. Stationary Points and Curve Sketching | 驻点与曲线草图

Stationary points occur where dy/dx = 0. To classify them as local maxima, minima or points of inflection, the second derivative test is usually fastest: if d²y/dx² > 0 the point is a minimum, if d²y/dx² < 0 it is a maximum, and if d²y/dx² = 0 you should check the sign of the gradient on either side.

驻点出现在 dy/dx = 0 的位置。要判断其为局部极大值、局部极小值或拐点，通常最快的方法是二阶导数判别法：若 d²y/dx² > 0 则为极小值点，若 d²y/dx² < 0 则为极大值点，若 d²y/dx² = 0 则应检查该点两侧梯度的符号。

y = x³ − 3x ⇒ dy/dx = 3x² − 3 = 3(x² − 1)

Setting dy/dx = 0 gives x = ±1. Since d²y/dx² = 6x, at x = 1 the second derivative is positive, so there is a minimum; at x = −1 it is negative, so there is a maximum. Curve sketching requires you to combine stationary points, intercepts and the behaviour as x → ±∞.

令 dy/dx = 0 可得 x = ±1。因为 d²y/dx² = 6x，在 x = 1 处二阶导数为正，因此存在极小值；在 x = −1 处二阶导数为负，因此存在极大值。画曲线草图时，需要综合驻点、截距以及 x → ±∞ 时的函数行为。

7. Introduction to Integration | 积分入门

Integration reverses differentiation. The indefinite integral of f(x) is written ∫ f(x) dx and must include an arbitrary constant +c because differentiation removes any constant term. For powers, the reverse power rule is used: add one to the power and divide by the new power, provided the original power is not −1.

积分是微分的逆运算。f(x) 的不定积分写作 ∫ f(x) dx，并且必须包含任意常数 +c，因为微分会消去任何常数项。对幂函数使用反向幂法则：先将指数加一，再除以新的指数，前提是原指数不为 −1。

∫ xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + c, n ≠ −1

For example, ∫ (4x³ − 2x + 5) dx = x⁴ − x² + 5x + c. You can always check an indefinite integral by differentiating the result to recover the original integrand, which is a useful exam technique.

例如，∫ (4x³ − 2x + 5) dx = x⁴ − x² + 5x + c。你总可以通过对结果求导来检验不定积分，看是否恢复原被积函数，这是考试中很实用的技巧。

8. Definite Integrals and Areas | 定积分与面积

A definite integral has limits a and b and is evaluated using the fundamental theorem of calculus: ∫ₐᵇ f(x) dx = F(b) − F(a), where F(x) is any antiderivative. Definite integrals give the signed area between the curve and the x-axis, so regions below the axis contribute negative values.

定积分带有上限 a 和下限 b，并使用微积分基本定理求值：∫ₐᵇ f(x) dx = F(b) − F(a)，其中 F(x) 是任意一个原函数。定积分给出曲线与 x 轴之间的有符号面积，因此位于 x 轴下方的区域贡献负值。

∫₁² 3x² dx = [x³]₁² = 8 − 1 = 7

To find the area between two curves, integrate the top function minus the bottom function over the required interval. You should also sketch the region and identify where the curves intersect, as this determines the limits of integration.

要求两条曲线之间的面积，应在所需区间上对上函数减去下函数进行积分。你还应画出区域草图并确定两曲线的交点，因为交点决定了积分上下限。

9. Reverse Chain Rule and Standard Integrals | 反链式法则与标准积分

Standard integrals must be memorised for Edexcel A Level Mathematics. They include ∫ eˣ dx = eˣ + c, ∫ 1/x dx = ln|x| + c, ∫ sin x dx = −cos x + c and ∫ cos x dx = sin x + c. These results appear constantly in both Pure and applied questions.

Edexcel A Level 数学要求熟记标准积分。它们包括 ∫ eˣ dx = eˣ + c、∫ 1/x dx = ln|x| + c、∫ sin x dx = −cos x + c 以及 ∫ cos x dx = sin x + c。这些结果在纯数学和应用题中经常出现。

The reverse chain rule is used when an integrand is of the form f'(g(x))g'(x). In that case the integral is simply f(g(x)) + c. This avoids writing a full substitution for simple linear inner functions.

当被积函数形如 f'(g(x))g'(x) 时，可使用反链式法则。此时积分结果就是 f(g(x)) + c。对于简单的线性内层函数，使用该方法可以避免完整的换元过程。

∫ (2x + 1)⁵ dx = (1/2) × (2x + 1)⁶ ÷ 6 + c = (2x + 1)⁶ ÷ 12 + c

Here the derivative of the inner function 2x + 1 is 2, so a factor of 1/2 is needed before integrating u⁵ with respect to u. The same idea works for exponential and trigonometric functions with linear inner arguments.

这里内层函数 2x + 1 的导数为 2，因此在对 u⁵ 关于 u 积分之前需要乘以 1/2。同样的思路适用于内层为线性函数的指数函数和三角函数。

10. Integration by Substitution | 换元积分法

Integration by substitution is one of the most powerful A2 techniques. Choose a substitution u = g(x), rewrite the integrand completely in terms of u, replace dx with du divided by du/dx, then integrate. For definite integrals, you must also change the limits.

换元积分法是 A2 阶段最强大的技巧之一。选择代换 u = g(x)，将被积函数完全改写为 u 的函数，将 dx 替换为 du 除以 du/dx，然后进行积分。对于定积分，还必须同时更换积分上下限。

For example, to find ∫ x√(x + 1) dx, let u = x + 1, so du/dx = 1 and x = u − 1. The integral becomes ∫ (u − 1)u¹/² du = ∫ (u³/² − u¹/²) du = (2/5)u⁵/² − (2/3)u³/² + c. Finally, substitute u = x + 1 back into the answer.

例如，要求 ∫ x√(x + 1) dx，令 u = x + 1，则 du/dx = 1，且 x = u − 1。积分变为 ∫ (u − 1)u¹/² du = ∫ (u³/² − u¹/²) du = (2/5)u⁵/² − (2/3)u³/² + c。最后将 u = x + 1 代回

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Calculus Essentials: Differentiation and Integration | 微积分核心：微分与积分