📚 AP Calculus: Top 100 Derivative and Integral Practice Questions on Key Concepts | AP 数学：微积分求导与积分100题考点精练
Mastering AP Calculus AB and BC requires deep familiarity with differentiation, integration, and the many ways these operations appear in applied problems. This article distills the most essential question types into a comprehensive review of 100 high-yield practice points. From basic power rules to advanced integration by parts and parametric motion, every concept is illustrated with representative examples and solving strategies that mirror the style and difficulty of the real exam.
掌握 AP 微积分 AB 与 BC 需要透彻理解求导、积分以及它们在应用题中的各种变形。本文把最核心的题型浓缩为 100 道高频考点的精练回顾。从基本幂函数法则到进阶的分部积分和参数运动，每个概念都用与真题风格和难度高度一致的例题和解题策略进行讲解。
1. Foundation: Limit Definition of the Derivative | 基础：导数的极限定义
The derivative as a limit is a recurring idea in multiple-choice questions. Be able to compute or interpret f'(x) = lim_{h→0} [f(x+h) – f(x)] / h. A classic trap is to test whether you recognize that a given limit represents the derivative at a specific point. For example, lim_{h→0} [sin(π+h) – sin(π)]/h = cos(π) = -1.
导数的极限定义在选择题中反复出现。必须能计算或解释 f'(x) = lim_{h→0} [f(x+h) – f(x)] / h。常见的陷阱是考查你能否识别某个给定的极限是否表示某一点的导数。例如 lim_{h→0} [sin(π+h) – sin(π)]/h = cos(π) = -1。
Practice task: Evaluate lim_{x→2} [ (x² – 4) / (x-2) ] using the derivative perspective. Immediately notice that it is the derivative of x² at x=2, which is 4.
练习任务：用导数的视角计算 lim_{x→2} [ (x² – 4) / (x-2) ]。立刻发现它正是 x² 在 x=2 处的导数，结果为 4。
2. Basic Differentiation Rules | 基本求导法则
Power rule: d/dx [xⁿ] = n xⁿ⁻¹. Sum, constant multiple, and product rules must be automatic. A typical question combines polynomial and exponential functions: find d/dx [ (3x² + 2x) eˣ ]. Apply the product rule to obtain (6x+2)eˣ + (3x²+2x)eˣ = eˣ (3x² + 8x + 2).
幂法则：d/dx [xⁿ] = n xⁿ⁻¹。和差、常数倍和乘法法则必须变成条件反射。典型题目结合多项式与指数函数：求 d/dx [ (3x² + 2x) eˣ ]。应用乘法法则得到 (6x+2)eˣ + (3x²+2x)eˣ = eˣ (3x² + 8x + 2)。
Quotient rule is useful but can often be avoided by rewriting. For y = (x²+1)/√x, rewrite as x^(3/2) + x^(-1/2) and differentiate term by term instead of invoking the quotient rule.
除法法则固然有用，但常可通过改写表达式来规避。对于 y = (x²+1)/√x，重写为 x^(3/2) + x^(-1/2) 再逐项求导，完全不需要动用除法法则。
3. Chain Rule Mastery | 链式法则精通
The chain rule appears in almost every derivative problem. Whether it’s a composition like sin(ln x) or e^(tan x), students must systematically differentiate the outer function and multiply by the derivative of the inner function. For complex nested functions, a layered approach helps: if y = cos³(√x), then y’ = 3 cos²(√x) * (-sin(√x)) * (1/(2√x)).
链式法则几乎出现在每一道求导题中。无论是 sin(ln x) 这样的复合，还是 e^(tan x)，都必须系统地对外层函数求导，再乘以内层函数的导数。对复杂的多层嵌套，分层处理效果极佳：若 y = cos³(√x)，则 y’ = 3 cos²(√x) · (-sin(√x)) · (1/(2√x))。
Implicit differentiation is essentially the chain rule applied to relations. Given x² + y² = 25, differentiate both sides with respect to x: 2x + 2y (dy/dx) = 0, so dy/dx = -x/y. Remember to treat y as a function of x.
隐函数求导本质上就是对关系式运用链式法则。给定 x² + y² = 25，两边对 x 求导：2x + 2y (dy/dx) = 0，故 dy/dx = -x/y。切记将 y 视为 x 的函数。
4. Derivatives of Transcendental Functions | 超越函数的导数
Exponential and logarithmic functions obey precise rules: d/dx [eˣ] = eˣ, d/dx [aˣ] = aˣ ln a, d/dx [ln x] = 1/x. Combine with the chain rule: d/dx [ln(sin x)] = cot x. For logarithmic differentiation, use ln y to simplify products or powers: y = xˣ → ln y = x ln x → y’/y = ln x + 1 → y’ = xˣ (ln x + 1).
指数函数和对数函数遵循精确的规则：d/dx [eˣ] = eˣ，d/dx [aˣ] = aˣ ln a，d/dx [ln x] = 1/x。与链式法则结合：d/dx [ln(sin x)] = cot x。对数求导法通过取 ln y 化简乘积和幂：y = xˣ → ln y = x ln x → y’/y = ln x + 1 → y’ = xˣ (ln x + 1)。
Trigonometric derivatives must be memorized: sin x → cos x, cos x → -sin x, tan x → sec² x, sec x → sec x tan x, csc x → -csc x cot x, cot x → -csc² x. Inverse trig derivatives are also tested: d/dx [arctan x] = 1/(1+x²).
三角函数导数必须熟记：sin x → cos x，cos x → -sin x，tan x → sec² x，sec x → sec x tan x，csc x → -csc x cot x，cot x → -csc² x。反三角函数导数也会考查：d/dx [arctan x] = 1/(1+x²)。
5. Higher-Order Derivatives and Taylor Polynomials | 高阶导数与泰勒多项式
The second derivative measures concavity and acceleration. Questions often ask to find f”(x) given f(x) = e^(2x) cos x. Use the product rule twice or organize your work carefully. The third or fourth derivative may appear in problems about Taylor polynomials: the coefficient of (x-a)ⁿ in the nth-degree Taylor polynomial is f⁽ⁿ⁾(a)/n!.
二阶导数衡量凹性和加速度。题目常要求已知 f(x) = e^(2x) cos x 求 f”(x)。必须两次运用乘法法则，或仔细组织计算过程。三阶、四阶导数可能出现在泰勒多项式问题中：n 次泰勒多项式中 (x-a)ⁿ 的系数为 f⁽ⁿ⁾(a)/n!。
Practice: Find the 3rd-degree Taylor polynomial for f(x) = ln x centered at x=1. Compute f(1)=0, f'(1)=1, f”(1)=-1, f”'(1)=2. Thus P₃(x) = (x-1) – (1/2)(x-1)² + (1/3)(x-1)³.
练习：求 f(x) = ln x 在 x=1 处的 3 次泰勒多项式。计算 f(1)=0, f'(1)=1, f”(1)=-1, f”'(1)=2。因此 P₃(x) = (x-1) – (1/2)(x-1)² + (1/3)(x-1)³。
6. Applications of Derivatives: Tangents and Linearization | 导数应用：切线与线性化
The equation of the tangent line at x=a is y = f(a) + f'(a)(x-a). Linear approximations use this to estimate function values. For example, to approximate √4.1, let f(x)=√x, a=4, f(4)=2, f'(4)=1/(2√4)=1/4, so √4.1 ≈ 2 + (1/4)(0.1) = 2.025. This connects directly to L’Hôpital’s rule when limits yield 0/0 or ∞/∞ forms; apply derivatives of numerator and denominator separately.
在 x=a 处的切线方程为 y = f(a) + f'(a)(x-a)。线性近似就是用它来估计函数值。例如，估算 √4.1，设 f(x)=√x, a=4, f(4)=2, f'(4)=1/(2√4)=1/4，因此 √4.1 ≈ 2 + (1/4)(0.1) = 2.025。这直接联系到洛必达法则，当极限呈现 0/0 或 ∞/∞ 型时，对分子分母分别求导即可。
7. Critical Points, Extrema, and Optimization | 临界点、极值与最优化
To find absolute extrema on a closed interval [a,b], evaluate f at critical points (where f’=0 or undefined) and at the endpoints. A typical optimization problem: A rectangle is inscribed under the parabola y = 12 – x² with its base on the x-axis. Express area A = 2x(12-x²) for x>0, then find A’ = 24 – 6x² = 0 → x=2, giving maximum area 32.
求闭区间 [a,b] 上的绝对极值，需评估 f 在临界点（f’=0 或不存在）以及端点处的值。典型的最优化问题：一个矩形内接于抛物线 y = 12 – x² 下方，底边在 x 轴上。面积 A = 2x(12-x²)，x>0，求导 A’ = 24 – 6x² = 0 → x=2，最大面积为 32。
Mean Value Theorem (MVT) and Extreme Value Theorem guarantee existence of certain points. MVT states there exists c in (a,b) such that f'(c) = (f(b)-f(a))/(b-a). Be ready to verify its conditions (continuous, differentiable) and solve for c.
中值定理（MVT）和极值定理保证了特定点的存在。MVT 表明在 (a,b) 内存在 c 使得 f'(c) = (f(b)-f(a))/(b-a)。要能验证其条件（连续、可导）并求解 c。
8. Riemann Sums and Definite Integrals | 黎曼和与定积分
The definite integral is defined as the limit of Riemann sums. Understand left, right, and midpoint approximations. Given a table of velocity values v(t) at t=0,2,4,6, approximate ∫₀⁶ v(t) dt using a left Riemann sum: Δt = 2, sum = 2[v(0)+v(2)+v(4)]. Interpretation: total distance travelled when velocity is positive.
定积分被定义为黎曼和的极限。需理解左、右、中点近似。已知速度表 v(t) 在 t=0,2,4,6 的值，用左黎曼和估算 ∫₀⁶ v(t) dt：Δt=2，和 = 2[v(0)+v(2)+v(4)]。物理解释：当速度为正值时，它代表总路程。
Trapezoidal rule: ∫ₐᵇ f(x) dx ≈ (Δx/2)[f(x₀)+2f(x₁)+2f(x₂)+…+f(xₙ)]. It often gives a better estimate than Riemann sums and is specifically tested in AP Calculus.
梯形法则：∫ₐᵇ f(x) dx ≈ (Δx/2)[f(x₀)+2f(x₁)+2f(x₂)+…+f(xₙ)]。它通常比黎曼和更精确，在 AP 微积分中会被专门考查。
9. Antiderivatives and Integration Techniques | 反导数与积分技巧
Master u-substitution: let u = inner function. For ∫ x sin(x²) dx, set u=x², du=2x dx → integral becomes (1/2)∫ sin u du = -(1/2) cos u + C = -(1/2) cos(x²) + C. Always adjust for du by constants. Recognize when a function and its derivative are paired: ∫ eˣ cos(eˣ) dx is a straightforward substitution.
精通换元积分法：设 u = 内层函数。对于 ∫ x sin(x²) dx，令 u=x²，du=2x dx → 积分变为 (1/2)∫ sin u du = -(1/2) cos u + C = -(1/2) cos(x²) + C。务必通过常数调整 du。学会辨认函数与其导数成对出现：∫ eˣ cos(eˣ) dx 就是一个直接的换元。
Integration by parts: ∫ u dv = uv – ∫ v du. Use LIATE (Log, Inverse trig, Algebraic, Trig, Exponential) to choose u. Example: ∫ x ln x dx → u=ln x, dv=x dx → du=(1/x)dx, v=x²/2. Then ∫ x ln x dx = (x²/2) ln x – ∫ (x²/2)(1/x) dx = (x²/2) ln x – x²/4 + C.
分部积分法：∫ u dv = uv – ∫ v du。使用 LIATE（对数、反三角、代数、三角、指数）顺序选择 u。例题：∫ x ln x dx → u=ln x, dv=x dx → du=(1/x)dx, v=x²/2。故 ∫ x ln x dx = (x²/2) ln x – ∫ (x²/2)(1/x) dx = (x²/2) ln x – x²/4 + C。
Partial fraction decomposition (BC): To integrate rational functions like (2x+3)/((x-1)(x+2)), write as A/(x-1) + B/(x+2), solve for A and B, then integrate to A ln|x-1| + B ln|x+2| + C.
部分分式分解（BC 考点）：对有理函数如 (2x+3)/((x-1)(x+2))，写为 A/(x-1) + B/(x+2)，解出 A 和 B，再积分得到 A ln|x-1| + B ln|x+2| + C。
10. Definite Integral Applications: Area, Volume, and Accumulation | 定积分应用：面积、体积与累积量
Area between curves: A = ∫ [top – bottom] dx or [right – left] dy. For y = x² and y = √x, intersection at x=0,1. Area = ∫₀¹ (√x – x²) dx = [ (2/3)x^(3/2) – (1/3)x³ ]₀¹ = 1/3. Volume by disc/washer: V = π ∫ (outer radius)² – (inner radius)² dx. Volume by known cross-sections: V = ∫ A(x) dx where A(x) is the area of a cross-section perpendicular to the x-axis.
曲线间面积：A = ∫ [上 – 下] dx 或 [右 – 左] dy。对于 y = x² 和 y = √x，交于 x=0,1。面积 = ∫₀¹ (√x – x²) dx = [ (2/3)x^(3/2) – (1/3)x³ ]₀¹ = 1/3。圆盘/垫圈法体积：V = π ∫ (外半径)² – (内半径)² dx。已知截面体积：V = ∫ A(x) dx，其中 A(x) 是垂直于 x 轴的截面面积。
Accumulation functions: F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x) by the Fundamental Theorem of Calculus Part 1. Be comfortable with chain rule in this setting: d/dx ∫₀^(x²) sin t dt = sin(x²) * 2x.
累积函数：F(x) = ∫ₐˣ f(t) dt，由微积分基本定理第一部分得 F'(x) = f(x)。要熟悉在此情境下应用链式法则：d/dx ∫₀^(x²) sin t dt = sin(x²) · 2x。
11. Differential Equations and Slope Fields | 微分方程与斜率场
Separable differential equations: dy/dx = g(x)h(y) → separate to (1/h(y)) dy = g(x) dx, then integrate both sides. Example: dy/dx = xy → (1/y) dy = x dx → ln|y| = x²/2 + C → y = Ce^(x²/2). Always solve for the constant when given an initial condition.
可分离微分方程：dy/dx = g(x)h(y) → 分离为 (1/h(y)) dy = g(x) dx，再对两边积分。例题：dy/dx = xy → (1/y) dy = x dx → ln|y| = x²/2 + C → y = Ce^(x²/2)。当给定初始条件时，务必解出常数。
Slope fields visually represent a differential equation by short line segments at grid points. Given a slope field, you may need to identify the matching differential equation or sketch a solution curve that passes through a given point.
斜率场通过在网格点上画出短线段来可视化微分方程。给定一个斜率场，可能需要找出匹配的微分方程，或者画出一条经过给定点的解曲线。
Exponential growth/decay: dy/dt = ky → y = y₀ e^(kt). Logistic growth (BC): dP/dt = kP(1 – P/L) with carrying capacity L. Recognize the form and interpret the solution curve.
指数增长与衰减：dy/dt = ky → y = y₀ e^(kt)。Logistic 增长（BC）：dP/dt = kP(1 – P/L)，L 为环境承载力。识别其形式并解释解曲线。
12. Parametric and Polar Calculus (BC Focus) | 参数与极坐标微积分（BC 重点）
For parametric equations x=f(t), y=g(t), the derivative dy/dx = (dy/dt)/(dx/dt). The second derivative d²y/dx² = d(dy/dx)/dt / (dx/dt). Arc length: ∫ √( (dx/dt)² + (dy/dt)² ) dt. Speed of a particle is the magnitude of velocity vector: √( [dx/dt]² + [dy/dt]² ).
对于参数方程 x=f(t), y=g(t)，导数 dy/dx = (dy/dt)/(dx/dt)。二阶导数 d²y/dx² = d(dy/dx)/dt / (dx/dt)。弧长：∫ √( (dx/dt)² + (dy/dt)² ) dt。粒子速率是速度向量的大小：√( [dx/dt]² + [dy/dt]² )。
Polar curves r = f(θ): area enclosed = (1/2) ∫ r² dθ. To find the slope of a polar curve at a point, convert to parametric: x = r cos θ, y = r sin θ, then compute dy/dx. Be prepared to find tangent lines at the pole or points where r=0.
极坐标曲线 r = f(θ)：围成面积 = (1/2) ∫ r² dθ。求极坐标曲线在某点的斜率，可转化为参数方程：x = r cos θ, y = r sin θ，再计算 dy/dx。准备好在极点或 r=0 处求切线。
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