📚 Exercise D.2: Further Differentiation | 练习 D.2：深入微分
Exercise D.2 in the IB Mathematics: Analysis and Approaches HL course is designed to move you beyond simple derivatives and into the full power of differential calculus. This set of problems typically demands fluency with the chain, product, and quotient rules as well as the ability to differentiate exponential, logarithmic, trigonometric, and implicitly defined functions. The following guide unpacks every essential technique, complete with worked examples and common pitfalls, so that you can approach your Exercise D.2 with real confidence.
IB 数学分析与方法 HL 课程中的练习 D.2 旨在让你超越简单的导数运算，全面掌握微分学的强大工具。这组题目通常要求熟练运用链式法则、乘积法则和商法则，并能对指数函数、对数函数、三角函数以及隐函数进行求导。下面的指南逐一讲解每一种核心技巧，并配有详细的例题和常见错误分析，帮助你充满信心地完成练习 D.2。
1. Recap of Basic Rules | 基本法则回顾
multiple rule, and sum rule at your fingertips. For any real power n, if f(x) = xⁿ then f'(x) = n xⁿ⁻¹. Constants factor out: d/dx [c·g(x)] = c·g'(x). Derivatives can be taken term by term: (u + v)' = u' + v' n by asking you to rewrite surds and fractions as powers of x before applying these rules.
在攻克更复杂的法则之前，你必须熟练掌握幂法则、常数倍法则以及和法则。对于任意实数指数 n，如果 f(x) = xⁿ，那么 f'(x) = n xⁿ⁻¹。常数可以提到微分符号外面：d/dx [c·g(x)] = c·g'(x)。导数可以逐项求导：(u + v)' = u' + v'。练习 D.2 中的许多题目都要求你先把根式和分式写成 x 的幂的形式，再运用这些法则。
d/dx (xⁿ) = n xⁿ⁻¹
d/dx (√x) = d/dx (x½) = ½ x⁻½
2. The Chain Rule | 链式法则
D.2. It tells us how to differentiate a composite function: if y = f(g(x)), then dy/dx = f'(g(x)) · g' if y = f(u) and u = g(x), then dy/dx = (dy/du) · (du/dx). Always work from the outside in – on, leaving the inner piece untouched, then multiply by the derivative of the inner piece.
链式法则可以说是练习 D.2 中使用频率最高的微分技巧。它告诉我们如何对复合函数求导：若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。用莱布尼兹记号表示就是，如果 y = f(u) 且 u = g(x)，那么 dy/dx = (dy/du) · (du/dx)。始终由外向内逐层求导——先对外层函数求导，保持内层部分不变，再乘以内层函数的导数。
Example: so y = u⁴. Then dy/du = 4u³, du/dx = 6x, giving dy/dx = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³.
示例：y = (3x² + 5)⁴。设 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x，得到 dy/dx = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³。
3. The Product Rule | 乘积法则
you cannot simply multiply the individual derivatives. The product rule states: dy/dx = u' v + u v' product rule with the chain rule, so be prepared to spot where an inner function appears.
当一个函数可以写成两个较简单函数的乘积，比如 y = u(x) v(x) 时，你不能简单地将各自的导数相乘。乘积法则告诉我们：dy/dx = u' v + u v'。用语言概括就是：先对第一个函数求导，保持第二个不变；加上保持第一个不变，对第二个求导。练习 D.2 中的很多题目都会把乘积法则与链式法则结合在一起使用，因此要随时准备识别出内层函数。
Example: y = x² · sin(3x). Let u = x², v = sin(3x). Then u' = 2x, v' by the chain rule. Hence dy/dx = 2x · sin(3x) + x² · 3 cos(3x) = 2x sin(3x) + 3x² cos(3x).
示例：y = x² · sin(3x)。设 u = x²，v = sin(3x)。那么 u' = 2x，而 v' 根据链式法则为 cos(3x) · 3。因此 dy/dx = 2x · sin(3x) + x² · 3 cos(3x) = 2x sin(3x) + 3x² cos(3x)。
4. The Quotient Rule | 商法则
v(x), where both u and v are differentiable and v(x) ≠ 0, the quotient rule is: dy/dx = (u' v – u v') / v². A common memory aid is “low d-high minus high d-low over low squared”. Be especially careful with the minus sign – r of x, consider rewriting the expression as a product and using the product rule instead.
对于分式形式的函数 y = u(x) / v(x)，只要 u 和 v 都可导且 v(x) ≠ 0，商法则就是：dy/dx = (u' v – u v') / v²。一个常用的记忆口诀是：“下面乘以上面的导数减去上面乘以下面的导数，再除以下面的平方。”要特别留意减号的位置——这是练习 D.2 中最常见的代数错误来源。如果分母只是一个 x 的幂，可以考虑把原式改写成乘积形式，再使用乘积法则。
Example: y = (x² + 1) / (x – 2). Let u = x² + 1, v = x – 2. Then u' = 2x, v' = 1. dy/dx = [2x(x – 2) – (x² + 1)(1)] / (x – 2)² = (2x² – 4x – x² – 1) / (x – 2)² = (x² – 4x – 1) / (x – 2)².
示例：y = (x² + 1) / (x – 2)。设 u = x² + 1，v = x – 2。则 u' = 2x，v' = 1。dy/dx = [2x(x – 2) – (x² + 1)(1)] / (x – 2)² = (2x² – 4x – x² – 1) / (x – 2)² = (x² – 4x – 1) / (x – 2)²。
5. Differentiating Exponential and Logarithmic Functions | 指数与对数函数微分
The natural exponential function eˣ is unique in that its derivative is itself: d/dx (eˣ) = eˣ. When the exponent is a function of x, apply the chain rule: d/dx (eᵍ⁽ˣ⁾) = eᵍ⁽ˣ⁾·g'(x). For the natural logarithm, d/dx (ln x) = 1/x, for x > 0. More generally, d/dx (ln |x|) = 1/x. The derivative of a general exponential aˣ is aˣ logarithms with an arbitrary base, use the change-of-base formula before differentiating.
自然指数函数 eˣ 的独特之处在于它的导数就是它本身：d/dx (eˣ) = eˣ。当指数部分是 x 的函数时，需要运用链式法则：d/dx (eᵍ⁽ˣ⁾) = eᵍ⁽ˣ⁾·g'(x)。对于自然对数，d/dx (ln x) = 1/x （x > 0）。更一般地，d/dx (ln |x|) = 1/x。一般指数函数 aˣ 的导数是 aˣ ln a。对于任意底数的对数，可以先使用换底公式再求导。
Example: y = e²ˣ⁺¹. d/dx (e²ˣ⁺¹) = e²ˣ⁺¹ · 2 = 2e²ˣ = 2/x, by first using log properties: ln(5x²) = ln 5 + 2 ln x, derivative = 0 + 2/x = 2/x.
示例：y = e²ˣ⁺¹。d/dx (e²ˣ⁺¹) = e²ˣ⁺¹ · 2 = 2e²ˣ⁺¹。y = ln(5x²)，可以先利用对数性质化为 ln 5 + 2 ln x，则导数为 0 + 2/x = 2/x；或用链式法则得到 (1/(5x²)) · 10x = 2/x。
6. Differentiating Trigonometric Functions | 三角函数的微分
he six trigonometric functions. The core ones are: d/dx (sin x) = cos x and d/dx (cos x) = – n derive the others using the quotient rule or chain rule: tan x gives sec² x, cot x gives –csc² x, sec x gives sec x tan x, and csc x gives – ians unless specified otherwise, and use the chain rule when the angle is a function of x.
练习 D.2 中的 IB 考题常常需要用到六个三角函数的导数。最核心的是：d/dx (sin x) = cos x 以及 d/dx (cos x) = –sin x。从这两个出发，借助商法则或链式法则可以推出其他函数的导数：tan x 的导数是 sec² x，cot x 的是 –csc² x，sec x 的是 sec x tan x，csc x 的是 –csc x cot x。记住，除非特别说明，角度一律以弧度为单位，且当角度是 x 的函数时务必使用链式法则。
Example: x² = 6x² cos(2x³). For y = tan(x), derivative is sec²(x). For θ in radians, d/dθ (cos θ) = –sin θ.
示例：y = sin(2x³)，d/dx [sin(2x³)] = cos(2x³) · 6x² = 6x² cos(2x³)。y = tan(x) 的导数为 sec²(x)。对于弧度制下的 θ，d/dθ (cos θ) = –sin θ。
7. Implicit Differentiation | 隐函数微分
on mixes x and y together, you need implicit differentiation. The approach is to differentiate both sides of the equation with respect to x, treating y as a function of x and therefore adding a dy/dx factor every time you differentiate a y-term. After differentiation, collect the dy/dx terms on one side and solve for dy/dx. Exercise D.2 often uses implicit differentiation to find tangents to curves like circles and elliptic shapes.
并非所有的关系都能写成显式的 y = f(x) 形式。当方程中混杂着 x 和 y 时，就需要隐函数微分。操作方法是对等式两边同时关于 x 求导，并将 y 视为 x 的函数，因此每次对含有 y 的项求导时都要乘上一个 dy/dx 因子。求导完毕后，将所有包含 dy/dx 的项移到等号同一边，然后解出 dy/dx。练习 D.2 中经常利用隐函数微分来求圆、椭圆等曲线的切线。
Example: Find dy/dx if x² + y² = 25. Differentiate: 2x + 2y·(dy/dx) = 0 → 2y (dy/dx) = –2x → dy/dx = –x/y. To find the tangent at (3,4), slope m = –3/4.
示例：已知 x² + y² = 25，求 dy/dx。两边求导：2x + 2y·(dy/dx) = 0 → 2y (dy/dx) = –2x → dy/dx = –x/y。在点 (3,4) 处的切线斜率 m = –3/4。
8. Logarithmic Differentiation | 对数微分法
entiation simplifies the process. You take the natural logarithm of both sides, use log properties to expand, differentiate implicitly, and then multiply through by the original y to obtain y'. This method is especially useful for functions of the form y = [f(x)]ᵍ⁽ˣ D.2 may include one or two problems that are far easier with logs than with direct rules.
当函数的底数和指数都含有变量，或者它是一个由多个因子组成的复杂乘积或商时，对数微分法可以大大简化求导过程。你可以对等式两边取自然对数，利用对数性质拆开，然后隐函数求导，最后两边同乘以原函数 y 得到 y'。这种方法对于形如 y = [f(x)]ᵍ⁽ˣ⁾ 的函数尤其有效。练习 D.2 中可能会有一两道题，使用对数处理远比直接求导轻松。
Example: y = xˣ iate implicitly: (1/y)(dy/dx) = 1·ln x + x·(1/x) = ln x + 1. Thus dy/dx = y (ln x + 1) = xˣ (ln x + 1).
示例：y = xˣ。取对数得 ln y = x ln x。隐函数求导：(1/y)(dy/dx) = ln x + 1。因此 dy/dx = y (ln x + 1) = xˣ (ln x + 1)。
9. Higher-Order Derivatives | 高阶导数
After finding the first derivative y' = f'(x), you may be asked in Exercise D.2 to compute the second derivative y'' = f'' ion in applications. Pay close attention to simplification after the first differentiation – into the second derivative often leads to errors. The notation d²y/dx² means d/dx(dy/dx).
在求得一阶导数 y' = f'(x) 后，练习 D.2 可能要求你计算二阶导数 y'' = f''(x)，甚至更高阶的导数。二阶导数就是一阶导数的导数，它在应用中描述了凹凸性和加速度。需要特别留意第一步求导后的化简——把一团乱麻带入第二次求导往往是错误的温床。符号 d²y/dx² 的含义是 d/dx(dy/dx)。
Example: y = x³ – 3x² + 2x. y' = 3x² – 6x + 2. y'' = 6x – 6. Setting y'' = 0 gives x = 1, a possible point of inflection.
示例：y = x³ – 3x² + 2x。y' = 3x² – 6x + 2，y'' = 6x – 6。令 y'' = 0 得 x = 1，这是一个可能的拐点。
10. Applications: Tangents and Normals | 应用：切线与法线
point. The tangent line has slope m = dy/dx evaluated at the point, and its equation is y – y₀ = m(x – x₀). The normal line is perpendicular to the tangent, so its slope is – ates correctly to find the y-intercept or write the equation in the requested linear form.
练习 D.2 中非常常见的一类问题是求给定点处的切线或法线方程。切线的斜率 m 等于该点处的 dy/dx 值，其方程为 y – y₀ = m(x – x₀)。法线与切线垂直，因此法线斜率为 –1/m（只要 m ≠ 0）。一定要准确代入点坐标，从而求出截距并按要求写成直线方程的形式。
Example: For y = x² + 1 at x = 1, y = 2. y' = 2x, so m_tangent = 2. Tangent: y – 2 = 2(x – 1) → y = 2x. Normal slope = –1/2, equation: y – 2 = –½(x – 1) → y = –½ x + 2.5.
示例：对于 y = x² + 1，当 x = 1 时 y = 2。y' = 2x，故切线斜率 m = 2。切线方程：y – 2 = 2(x – 1) → y = 2x。法线斜率为 –1/2，方程：y – 2 = –½(x – 1) → y = –½ x + 2.5。
11. Related Rates | 相关变化率
In related rates problems, two or more quantities are varying with time, and an equation relating them is known. Differentiating implicitly with respect to time t, using the chain rule to introduce d(quantity)/dt for each variable, allows you to solve for an unknown rate. Exercise D.2 may include geometrical situations – expanding circles, sliding ladders, or filling cones – where you must link radius, height, volume, and their time derivatives.
在相关变化率问题中，两个或多个量都随着时间变化，并且已知它们之间满足某个方程。通过对时间 t 进行隐函数求导，并利用链式法则为每一个变量引入 d(量)/dt，就可以解出未知的变化率。练习 D.2 中可能包含几何情境——例如不断扩大的圆、滑动的梯子或注入液体的圆锥——你必须将半径、高度、体积以及它们的时间导数联系起来。
Example: A spherical balloon' ge of volume V = (4/3)π r³ when r = 10 cm. dV/dt = 4π r² (dr/dt) = 4π·100·3 = 1200π cm³/s.
示例：一个气球的半径 r 以 3 cm/s 的速度增大。求当 r = 10 cm 时体积 V = (4/3)π r³ 的变化率。dV/dt = 4π r² (dr/dt) = 4π·100·3 = 1200π cm³/s。
12. Common Pitfalls and Tips | 常见误区与技巧
Exercise D.2 can feel overwhelming if you lose track of minus signs, forget the chain rule inside a product, or misapply the quotient rule. Always check if an expression can be simplified before differentiating – often rewriting √x or 1/x makes the work much cleaner. Double-check that the derivative of cos is –sin, not sin. When using implicit differentiation, never forget to attach dy/dx to y-terms. And finally, practise until the rules become automatic, because in the IB exam, Exercise D.2 skills underpin optimization and integration topics later.
如果弄丢了负号、在乘积内部忘了使用链式法则，或者用错了商法则，练习 D.2 确实会让人感到棘手。务必在求导前检查表达式能否化简——往往把 √x 或 1/x 改写成幂的形式会使计算清晰许多。反复确认 cos 的导数是 –sin 而不是 sin。进行隐函数求导时，千万别忘了对 y 项附上 dy/dx 因子。最后，要一直练习到法则运用自如，因为在 IB 考试中，练习 D.2 所涵盖的技能是后续优化问题和积分学习的基础。
Simplify first: Write radicals and denominators as powers of x.
先化简：将根式和分母写成 x 的幂。
Watch signs: The derivative of cos x is –sin x; the quotient rule has a minus sign.
注意符号：cos x 的导数是 –sin x；商法则中含有减号。
Chain rule inside: Don't forget to multiply by the derivative of the inner function.
内层链式：别忘了乘上内层函数的导数。
Implicit dy/dx: Every time you differentiate a y-term, attach dy/dx.
隐函数 dy/dx：每对含 y 的项求导，都要配上 dy/dx。
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