Exercise 21F: Mastering Differentiation Rules | 练习21F：掌握微分法则

📚 Exercise 21F: Mastering Differentiation Rules | 练习21F：掌握微分法则

Exercise 21F is designed to consolidate your understanding of the core differentiation rules at the heart of IB Mathematics: Analysis and Approaches. Whether you are working through the chain rule, product rule, or quotient rule, this exercise pushes you to combine techniques fluently and recognise when to apply each rule. The problems move from straightforward polynomials to composite, rational, and trigonometric functions, mirroring the progression you will see in Paper 1 and Paper 2. Mastering Exercise 21F means you can confidently differentiate almost any function encountered in the syllabus, and you will build the algebraic precision needed for optimisation, kinematics, and further calculus topics.

练习21F旨在巩固你对IB数学：分析与方法核心微分法则的理解。无论你正在练习链式法则、乘积法则还是商法则，这部分练习都要求你熟练地组合使用这些技巧，并能够判断何时应用哪条法则。题目从简单的多项式逐渐过渡到复合函数、有理函数和三角函数，这恰好反映了你在试卷一和试卷二中会遇到的难度进阶。掌握了练习21F，就意味着你能够自信地对课程中几乎任何函数进行求导，同时也为优化问题、运动学和后续微积分内容打下必要的代数精确度。

---

1. Reviewing the Power Rule and Basic Techniques | 复习幂法则与基本技巧

Before tackling the more complex combinations, Exercise 21F begins by reinforcing the power rule and the differentiation of simple trigonometric functions. For any real number n, the derivative of xⁿ is n xⁿ⁻¹. This rule extends naturally to constant multiples and sums: if f(x) = a u(x) + b v(x), then f'(x) = a u'(x) + b v'(x). You are also expected to recall the derivatives of sin x, cos x, and tan x, which are cos x, −sin x, and sec² x respectively. These fundamentals form the bedrock upon which the product, quotient, and chain rules are built.

在挑战更复杂的组合之前，练习21F首先强化幂法则以及简单三角函数的求导。对于任意实数n，xⁿ的导数是n xⁿ⁻¹。这一法则自然地推广到常数倍和和差运算：若f(x) = a u(x) + b v(x)，则f'(x) = a u'(x) + b v'(x)。你还应牢记sin x、cos x和tan x的导数，分别为cos x、−sin x和sec² x。这些基础内容是构建乘积法则、商法则和链式法则的基石。

Consider the function f(x) = 3x⁵ − 4x³ + 2x − 7. Applying the power rule term by term gives f'(x) = 15x⁴ − 12x² + 2. For a trigonometric example, g(x) = 2 sin x − 5 cos x differentiates to g'(x) = 2 cos x + 5 sin x. Notice how the sign changes carefully: the derivative of −5 cos x is +5 sin x because the derivative of cos x is −sin x. These straightforward applications appear early in Exercise 21F to warm you up for the multi-step problems ahead.

考虑函数f(x) = 3x⁵ − 4x³ + 2x − 7。逐项应用幂法则可得f'(x) = 15x⁴ − 12x² + 2。对于三角函数示例，g(x) = 2 sin x − 5 cos x的导数为g'(x) = 2 cos x + 5 sin x。请特别注意符号的变化：−5 cos x的导数为+5 sin x，这是因为cos x的导数是−sin x。这些直接的应用出现在练习21F的前半部分，为你预热，以便应对后续需要多个步骤的题目。

---

2. The Product Rule in Action | 乘积法则的实际运用

When a function is expressed as the product of two differentiable functions, u(x) and v(x), we use the product rule: (u v)’ = u’ v + u v’. The IB syllabus expects you to state this rule and apply it to algebraic and trigonometric combinations. A common mistake is to assume the derivative of a product is simply the product of the derivatives—Exercise 21F deliberately includes counterexamples to dispel this misconception. You must learn to identify the two factors, differentiate each separately, and then assemble the result.

当一个函数可以表示为两个可微函数的乘积u(x)和v(x)时，我们使用乘积法则：(u v)’ = u’ v + u v’。IB课程要求你能够表述这一法则，并将其应用于代数与三角函数的组合中。一个常见的错误是认为乘积的导数就等于导数的乘积——练习21F特意包含了反例来纠正这一误解。你必须学会识别两个因子，分别求导，然后再组合出最终结果。

For instance, to differentiate f(x) = x² sin x, let u = x² and v = sin x. Then u’ = 2x and v’ = cos x. The product rule gives f'(x) = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x. In some cases, you may need to factor the final answer; IB markschemes often accept either a factorised or an expanded form, but a neat, simplified expression is always preferred. Exercise 21F includes problems where you must also apply the chain rule within one of the factors, such as h(x) = eˣ cos(2x), preparing you for the interplay of multiple rules.

例如，要对f(x) = x² sin x求导，设u = x²，v = sin x。则u’ = 2x，v’ = cos x。乘积法则给出f'(x) = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x。在某些情形下，你可能需要对最终答案进行因式分解；IB的评分标准通常接受因式分解形式或展开形式，但一个整洁、化简后的表达式总是更受青睐。练习21F中还包括了一些需要你在某个因子内部同时使用链式法则的题目，比如h(x) = eˣ cos(2x)，这为你应对多重法则的结合做好了准备。

---

3. The Quotient Rule: Structure and Sign Awareness | 商法则：结构与符号意识

The quotient rule handles functions of the form f(x) = u(x) / v(x). The formula is (u/v)’ = (u’ v − u v’) / v². Order matters critically: the numerator begins with u’ v, not v u’. This often causes sign errors, especially when trigonometric or exponential functions are involved. Exercise 21F builds your fluency by gradually increasing the complexity of the numerator and denominator, from simple polynomials to rational trigonometric expressions.

商法则处理形如f(x) = u(x) / v(x)的函数。其公式为(u/v)’ = (u’ v − u v’) / v²。这里的顺序至关重要：分子以u’ v开头，而不是v u’。这常常导致符号错误，尤其当涉及三角函数或指数函数时。练习21F通过逐步提升分子与分母的复杂程度——从简单的多项式到有理三角表达式——来增强你的熟练度。

Take f(x) = x / cos x. Here u = x, v = cos x, so u’ = 1, v’ = −sin x. Applying the quotient rule yields f'(x) = (1·cos x − x·(−sin x)) / cos² x = (cos x + x sin x) / cos² x. A common pitfall is forgetting to square the denominator or misplacing the minus sign. Another typical problem from Exercise 21F is differentiating a rational function like f(x) = (x² + 1) / (x³ − 2). Here u’ = 2x, v’ = 3x², giving f'(x) = (2x(x³ − 2) − (x² + 1)·3x²) / (x³ − 2)². Simplify the numerator algebraically: 2x⁴ − 4x − 3x⁴ − 3x² = −x⁴ − 3x² − 4x, so the final derivative is (−x⁴ − 3x² − 4x) / (x³ − 2)².

以f(x) = x / cos x为例。这里u = x，v = cos x，因此u’ = 1，v’ = −sin x。应用商法则得到f'(x) = (1·cos x − x·(−sin x)) / cos² x = (cos x + x sin x) / cos² x。一个常见陷阱是忘记将分母平方，或者写错负号。练习21F中另一类典型题目是对有理函数的求导，例如f(x) = (x² + 1) / (x³ − 2)。这里u’ = 2x，v’ = 3x²，得到f'(x) = (2x(x³ − 2) − (x² + 1)·3x²) / (x³ − 2)²。代数化简分子：2x⁴ − 4x − 3x⁴ − 3x² = −x⁴ − 3x² − 4x，所以最终导数为(−x⁴ − 3x² − 4x) / (x³ − 2)²。

---

4. The Chain Rule: Differentiating Composite Functions | 链式法则：复合函数的求导

The chain rule is arguably the most powerful differentiation tool in the IB syllabus, and Exercise 21F dedicates substantial space to its mastery. If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In Leibniz notation, if y = f(u) and u = g(x), then dy/dx = (dy/du) · (du/dx). This rule allows you to differentiate functions like √(x² + 1), e^(sin x), or ln(cos x) by peeling back the layers from the outside in. The key is to identify the inner function correctly and to continue differentiating until you reach the bare variable.

链式法则可以说是IB课程中最强大的求导工具，练习21F用了相当大的篇幅来帮助你掌握它。若y = f(g(x))，则dy/dx = f'(g(x)) · g'(x)。用莱布尼茨记号表示，若y = f(u)且u = g(x)，则dy/dx = (dy/du) · (du/dx)。这条法则使你能够对诸如√(x² + 1)、e^(sin x)或ln(cos x)这样的函数求导，方法是从外向内一层层剥离。关键在于正确识别内层函数，并持续求导直到只剩下变量本身。

For y = (3x² − 5)⁴, the outer function is u⁴ and the inner function is u = 3x² − 5. Thus dy/dx = 4(3x² − 5)³ · 6x = 24x(3x² − 5)³. For trigonometric composites like y = sin(2x + 1), the derivative is cos(2x + 1) · 2 = 2 cos(2x + 1). Exercise 21F often combines the chain rule with the product or quotient rule. For example, to differentiate y = x²(2x + 1)³, you might use the product rule with u = x² and v = (2x + 1)³, and then apply the chain rule to find v’ = 3(2x + 1)²·2 = 6(2x + 1)². The complete derivative is 2x(2x + 1)³ + x²·6(2x + 1)², which simplifies to 2x(2x + 1)²[(2x + 1) + 3x] = 2x(2x + 1)²(5x + 1).

对于y = (3x² − 5)⁴，外层函数是u⁴，内层函数是u = 3x² − 5。因此dy/dx = 4(3x² − 5)³ · 6x = 24x(3x² − 5)³。对于像y = sin(2x + 1)这样的三角复合函数，其导数为cos(2x + 1) · 2 = 2 cos(2x + 1)。练习21F经常将链式法则与乘积法则或商法则结合使用。例如，要对y = x²(2x + 1)³求导，你可以先使用乘积法则，令u = x²、v = (2x + 1)³，然后应用链式法则求得v’ = 3(2x + 1)²·2 = 6(2x + 1)²。完整的导数为2x(2x + 1)³ + x²·6(2x + 1)²，化简后得到2x(2x + 1)²[(2x + 1) + 3x] = 2x(2x + 1)²(5x + 1)。

---

5. Combining the Rules: Strategy and Order | 法则的联合使用：策略与顺序

Many problems in Exercise 21F require you to use two or even three rules in a single differentiation. The order in which you apply them is determined by the structure of the function’s expression. As a general guideline, examine the outermost operation: if the function is essentially a product of two blocks, use the product rule first; if it is a quotient, use the quotient rule; if it is a composition (function inside a function), start with the chain rule. Inside each block, you may then need further rules. Writing out a clear plan before you begin differentiating can prevent algebraic chaos.

练习21F中的许多题目要求你在一次求导中使用两个甚至三个法则。你应用这些法则的顺序取决于函数表达式的结构。作为一般性指导原则，请检查最外层的运算：如果函数本质上是两个块的乘积，先使用乘积法则；如果是一个商，则使用商法则；如果是复合（函数内嵌函数），则从链式法则开始。在每个块的内部，你可能还需要进一步的法则。在动手求导之前写出清晰的计划，可以避免代数上的混乱。

Let’s differentiate f(x) = (e^(3x) · ln x) / (x² + 1). The outermost structure is a quotient, so we label u = e^(3x) · ln x and v = x² + 1. To find u’, we need the product rule on e^(3x) and ln x, and the chain rule for e^(3x). Specifically, derivative of e^(3x) is 3e^(3x), so u’ = 3e^(3x) · ln x + e^(3x) · (1/x). For v’ we simply have 2x. Now apply the quotient rule: f'(x) = [u’·v − u·v’] / v². Substituting gives a complex but manageable rational expression that can be simplified by factoring e^(3x). This layered approach is typical of the later parts of Exercise 21F.

我们来对f(x) = (e^(3x) · ln x) / (x² + 1)进行求导。最外层的结构是一个商，因此我们标记u = e^(3x) · ln x和v = x² + 1。为了求u’，我们需要对e^(3x)和ln x使用乘积法则，并对e^(3x)使用链式法则。具体来说，e^(3x)的导数是3e^(3x)，因此u’ = 3e^(3x) · ln x + e^(3x) · (1/x)。对于v’，我们直接得到2x。现在应用商法则：f'(x) = [u’·v − u·v’] / v²。代入后得到一个复杂但可处理的分式，可以通过提取公因式e^(3x)进行化简。这种分层处理的方法是练习21F后半部分的典型特征。

---

6. Differentiating Exponential and Logarithmic Functions | 指数函数与对数函数的求导

IB Mathematics places a strong emphasis on the natural exponential function eˣ and the natural logarithm ln x. Recall that d/dx (eˣ) = eˣ, a beautifully self-reproducing property. When the exponent is a function of x, the chain rule gives d/dx (e^(g(x))) = e^(g(x)) · g'(x). For logarithms, d/dx (ln x) = 1/x, and more generally, d/dx (ln(g(x))) = g'(x) / g(x). Exercise 21F contains multiple variations that test your ability to differentiate aˣ (where a is a constant) using the conversion aˣ = e^(x ln a), a favourite exam trick.

IB数学非常重视自然指数函数eˣ和自然对数ln x。请记住d/dx (eˣ) = eˣ，这是一个优美的自复制性质。当指数是x的函数时，链式法则给出d/dx (e^(g(x))) = e^(g(x)) · g'(x)。对于对数，d/dx (ln x) = 1/x，更一般地，d/dx (ln(g(x))) = g'(x) / g(x)。练习21F包含了多种变式，用来检验你对形如aˣ（a为常数）的函数进行求导的能力——你需要利用转换aˣ = e^(x ln a)，这是考试中常见的技巧。

Consider f(x) = 2ˣ. Rewrite as f(x) = e^(x ln 2), so f'(x) = e^(x ln 2) · ln 2 = 2ˣ ln 2. Another classic problem is f(x) = ln(3x² − 2x). Using the chain rule, f'(x) = (6x − 2) / (3x² − 2x). You can often simplify by factoring: (2(3x − 1)) / (x(3x − 2)). When the argument of the logarithm involves a product, quotient, or power, it is sometimes easier to apply logarithmic properties before differentiating. For example, y = ln( (x² sin x) / eˣ ) can be expanded to ln(x²) + ln(sin x) − ln(eˣ) = 2 ln x + ln(sin x) − x, which differentiates much more easily to 2/x + cot x − 1.

考虑f(x) = 2ˣ。将其改写为f(x) = e^(x ln 2)，因此f'(x) = e^(x ln 2) · ln 2 = 2ˣ ln 2。另一道经典题目是f(x) = ln(3x² − 2x)。运用链式法则，f'(x) = (6x − 2) / (3x² − 2x)。你通常可以通过因式分解进行化简：(2(3x − 1)) / (x(3x − 2))。当对数的真数包含乘积、商或幂时，有时先在求导前运用对数性质进行展开会更简单。例如，y = ln( (x² sin x) / eˣ )可以展开为ln(x²) + ln(sin x) − ln(eˣ) = 2 ln x + ln(sin x) − x，这样求导起来就简单得多，结果是2/x + cot x − 1。

---

7. Trigonometric Functions and Repeated Differentiation | 三角函数与反复求导

Trigonometric differentiation extends well beyond the basic sine and cosine rules. Exercise 21F ensures you are comfortable with the derivatives of tan x, sec x, csc x, and cot x. It is worth memorising that d/dx (tan x) = sec² x and d/dx (sec x) = sec x tan x. The reciprocal functions (cot, csc) follow similar patterns but with negative signs. Moreover, the chain rule frequently appears in tandem with trig functions: d/dx (sin(ax + b)) = a cos(ax + b), and d/dx (cos² x) = 2 cos x · (−sin x) = −2 sin x cos x = −sin 2x. Recognising these simplifications can save time and align with IB’s emphasis on exact values and neat forms.

三角函数的求导远不止基本的正弦和余弦法则。练习21F确保你熟练掌握tan x、sec x、csc x和cot x的导数。值得记忆的是d/dx (tan x) = sec² x以及d/dx (sec x) = sec x tan x。倒数函数（cot, csc）遵循类似的模式，但带有负号。此外，链式法则经常与三角函数一同出现：d/dx (sin(ax + b)) = a cos(ax + b)，而d/dx (cos² x) = 2 cos x · (−sin x) = −2 sin x cos x = −sin 2x。识别这些化简方式可以节省时间，也符合IB对精确值和整洁形式的重视。

Higher-order derivatives (second derivative, third derivative) also appear in Exercise 21F, often in the context of kinematics or curve sketching. If s(t) represents displacement, then velocity v(t) = s'(t) and acceleration a(t) = s”(t). For a trigonometric motion like s(t) = 3 sin(π t), we have v(t) = 3π cos(π t) and a(t) = −3π² sin(π t). Repeated differentiation of simple trig functions leads to cyclic patterns: differentiating sin x twice brings you back to −sin x. This cyclic property is sometimes tested in more abstract problems within the exercise.

高阶导数（二阶导数、三阶导数）同样出现在练习21F中，通常是在运动学或曲线作图的背景下。若s(t)表示位移，那么速度v(t) = s'(t)而加速度a(t) = s”(t)。对于像s(t) = 3 sin(π t)这样的三角运动，我们有v(t) = 3π cos(π t)以及a(t) = −3π² sin(π t)。对简单三角函数进行反复求导会产生循环模式：对sin x求导两次会回到−sin x。这种循环性质有时会在练习中更抽象的题目中考察。

---

8. Implicit Differentiation Basics | 隐函数求导基础

While Exercise 21F primarily focuses on explicit differentiation, some IB courses introduce the first steps of implicit differentiation here. When an equation relates x and y without explicitly solving for y, we can still find dy/dx by differentiating both sides with respect to x, treating y as a function of x. Whenever you differentiate a term involving y, you must multiply by dy/dx (the chain rule). This technique is particularly useful for circles, ellipses, and other curves where isolating y is messy or impossible.

虽然练习21F主要聚焦于显函数求导，但一些IB课程会在此处引入隐函数求导的初步内容。当一个方程关联了x和y，却并未明确解出y时，我们仍然可以通过对等式两边关于x求导来找到dy/dx，此时应将y视为x的函数。每当你对含有y的项求导时，必须乘以dy/dx（链式法则）。这一技巧对于圆、椭圆以及其他难以或无法解出y的曲线尤其有用。

For a simple circle x² + y² = 25, differentiate termwise: 2x + 2y · dy/dx = 0. Solving gives dy/dx = −x/y. If the equation includes products like x y³, use the product rule: d/dx (x y³) = 1·y³ + x·3y²·dy/dx. Implicit differentiation questions in Exercise 21F typically ask you to find the slope of the tangent at a given point, so you will substitute the coordinates after finding the derivative expression. This skill bridges differentiation with coordinate geometry and sets the stage for related rates problems.

对于简单的圆x² + y² = 25，逐项求导：2x + 2y · dy/dx = 0。解出dy/dx = −x/y。如果方程中包含像x y³这样的乘积，则使用乘积法则：d/dx (x y³) = 1·y³ + x·3y²·dy/dx。练习21F中的隐函数求导题目通常会要求你找出在给定点处切线的斜率，因此你需要在得到导数表达式后代入坐标值。这一技巧将微分与坐标几何联系起来，并为相关变化率问题奠定了基础。

---

9. Common Errors and How to Avoid Them | 常见错误及其避免方法

Even strong students lose marks on Exercise 21F-style problems due to predictable errors. The most frequent mistake is misapplying the quotient rule by inverting the terms in the numerator (writing v u’ − u v’ instead of u’ v − u v’). Another is forgetting the chain rule entirely, such as writing the derivative of sin(2x) as cos(2x) without the factor of 2. Similarly, when differentiating e^(−x), the derivative is −e^(−x), not e^(−x). Being meticulous with signs is non-negotiable.

即便是能力较强的学生，在类似练习21F的题目中也会因为一些可以预见的错误而丢分。最常见的错误是在使用商法则时颠倒了分子中的项（写成了v u’ − u v’，而不是u’ v − u v’）。另一个错误是完全忘记链式法则，比如把sin(2x)的导数写成cos(2x)而漏掉了因子2。类似地，在对e^(−x)求导时，导数是−e^(−x)，而不是e^(−x)。对符号的仔细处理是毫无商量余地的。

Algebraic simplification errors also creep in when students combine rules. After applying the product or quotient rule, always check if the numerator can be factored. Factoring often reveals cancellations or simplifications that reduce the expression to a more manageable form. Additionally, misidentifying the inner function in a chain rule application can derail the entire problem. Always ask: “What is the last operation being applied to x?” and let that guide your choice of u. Writing out u, u’, v, v’ clearly on your paper before assembling the final formula is a simple habit that dramatically reduces errors.

代数化简错误也常在学生综合运用法则时悄悄出现。在应用乘积法则或商法则之后，务必检查分子是否可以因式分解。因式分解常常能揭示出可以约分或化简的部分，将表达式化至更易处理的形式。此外，在应用链式法则时错误地识别内层函数也会导致整个问题偏离轨道。要始终问自己：“对x进行的最后一步运算是什么？”让这个问题的答案引导你选择u。在组装最终公式前，在草稿纸上清楚地写出u、u’、v、v’，这是一个简单却可以大幅减少错误的习惯。

---

10. Connecting Exercise 21F to IB Exam Success | 将练习21F与IB考试成功联系起来

Exercise 21F is not an isolated set of drills; it is a microcosm of the differentiation component on your IB Mathematics exams. In Paper 1 (non-calculator), you may be asked to find f'(x) for a composite function and then evaluate it at a specific point, requiring exact values and often involving π. In Paper 2, differentiation questions can be embedded in larger problems about graph behaviour, optimisation, or motion. The algebraic fluency you gain from thoroughly working through Exercise 21F will directly translate to faster, more accurate solutions under time pressure.

练习21F并非一组孤立的训练题；它是IB数学考试中微分部分的一个缩影。在试卷一（不可使用计算器）中，你可能会被要求找出某个复合函数的f'(x)，然后计算其在某一点的值，这需要精确值并常涉及π。在试卷二中，求导题目可能会嵌套在关于图像形态、优化或运动的更大型问题中。你通过彻底完成练习21F所学得的代数熟练度，将直接转化为在时间压力下更快、更准确的解题能力。

Additionally, the ability to differentiate correctly underpins the entire calculus option and topics like Maclaurin series and differential equations in the HL syllabus. If you are aiming for a level 7, every mark counts, and a silly slip in a derivative can cascade into a cascade of errors in integration or area problems. Treat Exercise 21F as a diagnostic tool: if you can complete every problem without hesitation, you are truly ready for the exam. If certain rule combinations still feel awkward, isolate those patterns and practice similar examples until the process becomes automatic.

此外，正确求导的能力是整个微积分选修内容以及IB HL课程中麦克劳林级数和微分方程等专题的基础。如果你志在取得7分，那么每一分都至关重要，在导数部分一个愚蠢的笔误可能会在积分或面积问题中引发一连串错误。请将练习21F视为一项诊断工具：如果你能毫不犹豫地完成每一道题目，那么你就真正为考试做好了准备。如果某些法则组合仍让你感到别扭，请将它们单独拎出来，并针对类似示例反复练习，直至整个过程变得自动自如。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）