📚 Exercise 22C.1: Differentiation of Composite and Product Functions | 习题22C.1：复合函数与乘积函数的求导
In IB Mathematics, differentiation stands as a core analytical tool. Exercise 22C.1 presents a function that combines both product and composite elements: y = (2x+1)³(3x-2)⁴. This article dissects the problem step by step, reinforces the underlying rules, and extends into practical applications such as tangent equations and optimization, equipping you with a robust differentiation toolkit.
在IB数学中，求导是一项核心的分析工具。习题22C.1给出一个综合了乘积与复合结构的函数：y = (2x+1)³(3x-2)⁴。本文将逐步剖析此题，巩固基本法则，并延展至切线方程、优化等实际应用，为你构建一套扎实的求导工具箱。
1. Understanding the Problem | 理解题目
The target function is y = (2x+1)³(3x-2)⁴. We immediately see a product of two expressions, each raised to a power. To differentiate correctly, we need to combine the product rule and the chain rule. Identify the first factor u = (2x+1)³ and the second factor v = (3x-2)⁴.
目标函数是 y = (2x+1)³(3x-2)⁴。我们立刻看出它是两个各自带有幂次的表达式之积。要正确求导，必须联合运用乘积法则与链式法则。先将第一因式定为 u = (2x+1)³，第二因式定为 v = (3x-2)⁴。
2. The Product Rule | 乘积法则
When y = u v, the derivative is given by dy/dx = u’ v + u v’. This rule is indispensable for products. A handy mnemonic is “derivative of the first times the second, plus the first times the derivative of the second.” Always apply it before simplifying.
当 y = u v 时，导数公式为 dy/dx = u’ v + u v’。处理乘积函数时，此法则不可或缺。一个简便记法是“前导乘后，加前乘后导”。务必在化简前正确使用它。
| Product Rule | (u v)’ = u’ v + u v’ |
在IB数学考试中，乘积法则是高频考点，稍有不慎便会漏项或错序。
3. The Chain Rule | 链式法则
Each factor u = (2x+1)³ and v = (3x-2)⁴ is a composite function. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) × g'(x). For a power of a linear function, multiply by the derivative of the inner linear term.
每个因式 u = (2x+1)³ 和 v = (3x-2)⁴ 都是复合函数。链式法则指出，若 y = f(g(x))，则 dy/dx = f'(g(x)) × g'(x)。对于线性函数的幂次，需乘以内层线性项的导数。
| Chain Rule | d/dx [f(g(x))] = f'(g(x)) · g'(x) |
链式法则的本质是“外层求导乘以内层导数”，忘记内层乘子是常见错误。
4. Step-by-Step Derivation | 逐步求导过程
Step 1: Differentiate u = (2x+1)³ using the chain rule. Let inner function w = 2x+1, so u = w³. Then du/dx = 3w² × dw/dx = 3(2x+1)² × 2 = 6(2x+1)².
步骤1：运用链式法则对 u = (2x+1)³ 求导。设内层函数 w = 2x+1，则 u = w³。于是 du/dx = 3w² × dw/dx = 3(2x+1)² × 2 = 6(2x+1)²。
Step 2: Differentiate v = (3x-2)⁴. Let inner function z = 3x-2, so v = z⁴. Then dv/dx = 4z³ × dz/dx = 4(3x-2)³ × 3 = 12(3x-2)³.
步骤2：对 v = (3x-2)⁴ 求导。设内层函数 z = 3x-2，则 v = z⁴。因此 dv/dx = 4z³ × dz/dx = 4(3x-2)³ × 3 = 12(3x-2)³。
Step 3: Assemble using the product rule: dy/dx = u’ v + u v’ = 6(2x+1)²(3x-2)⁴ + (2x+1)³ × 12(3x-2)³ = 6(2x+1)²(3x-2)⁴ + 12(2x+1)³(3x-2)³.
步骤3：用乘积法则整合：dy/dx = u’ v + u v’ = 6(2x+1)²(3x-2)⁴ + (2x+1)³ × 12(3x-2)³ = 6(2x+1)²(3x-2)⁴ + 12(2x+1)³(3x-2)³。
This is a correct but unsimplified derivative. We can leave it here or simplify further, which is often rewarded in IB marking schemes.
这是一个正确但未化简的导数。可以停留在此，或进一步化简，后者往往在IB评分中获得认可。
5. Simplifying the Derivative | 化简导数
Factor out the greatest common factor: 6(2x+1)²(3x-2)³. Then dy/dx = 6(2x+1)²(3x-2)³ [ (3x-2) + 2(2x+1) ]. Simplify inside brackets: (3x-2) + (4x+2) = 7x. Finally, dy/dx = 6(2x+1)²(3x-2)³ × 7x = 42x(2x+1)²(3x-2)³.
提取最大公因式：6(2x+1)²(3x-2)³。则 dy/dx = 6(2x+1)²(3x-2)³ [ (3x-2) + 2(2x+1) ]。化简括号内：(3x-2) + (4x+2) = 7x。最终，dy/dx = 6(2x+1)²(3x-2)³ × 7x = 42x(2x+1)²(3x-2)³。
dy/dx = 42x (2x+1)² (3x-2)³
This compact form is easy to evaluate at any point and clearly reveals the critical point at x = 0.
这一紧凑形式便于在任意点求值，并清晰显示出 x = 0 处的驻点。
6. Verification Using Logarithmic Differentiation | 使用对数求导法验证
Take natural logs of both sides: ln y = 3 ln(2x+1) + 4 ln(3x-2). Differentiate implicitly: (1/y) dy/dx = 3 × (2/(2x+1)) + 4 × (3/(3x-2)) = 6/(2x+1) + 12/(3x-2). Thus dy/dx = y [6/(2x+1) + 12/(3x-2)]. Multiply y back: (2x+1)³(3x-2)⁴ [6/(2x+1) + 12/(3x-2)] = (2x+1)²(3x-2)³ [6(3x-2) + 12(2x+1)] = 6(2x+1)²(3x-2)³ (7x) = 42x(2x+1)²(3x-2)³. It matches, confirming our result.
对等式两边取自然对数：ln y = 3 ln(2x+1) + 4 ln(3x-2)。隐式求导：(1/y) dy/dx = 3 × (2/(2x+1)) + 4 × (3/(3x-2)) = 6/(2x+1) + 12/(3x-2)。因此 dy/dx = y [6/(2x+1) + 12/(3x-2)]。将 y 乘回：(2x+1)³(3x-2)⁴ [6/(2x+1) + 12/(3x-2)] = (2x+1)²(3x-2)³ [6(3x-2) + 12(2x+1)] = 6(2x+1)²(3x-2)³ (7x) = 42x(2x+1)²(3x-2)³。结果吻合，验证无误。
Logarithmic differentiation is especially powerful when both base and exponent are functions, but it also serves as an excellent check for product-heavy expressions.
对数求导法在处理底数和指数均为函数时尤为强大，但对多重乘积表达式，它也是一个极好的检验工具。
7. Common Mistakes to Avoid | 常见错误
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Forgetting the inner derivative when applying the chain rule to (2x+1)³ or (3x-2)⁴. Always multiply by the derivative of the linear term.
对 (2x+1)³ 或 (3x-2)⁴ 应用链式法则时忘记内层导数。切记乘上线性项的导数。
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Misapplying the product rule by only differentiating one factor. Both u’v and uv’ must be included.
错误使用乘积法则，只对其中一个因子求导。必须包含 u’v 和 uv’ 两项。
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Incorrect simplification, such as losing the factor x when factoring, or arithmetic slips when expanding brackets.
化简错误，比如提取公因式时丢失因子 x，或展开括号时出现算术疏忽。
Double-checking each step with an alternative method or by substituting a simple value can catch most errors.
用替代方法或代入简单数值验算每个步骤，可以拦截大多数错误。
8. Application: Finding the Equation of a Tangent | 应用：求切线方程
Suppose we want the tangent to the curve at x = 1. First compute y: y = (2×1+1)³(3×1-2)⁴ = (3)³(1)⁴ = 27. Then compute dy/dx using the simplified form: dy/dx = 42×1×(3)²×(1)³ = 42×9 = 378. The tangent line has slope 378 and passes through (1, 27). Equation: y – 27 = 378(x – 1), or y = 378x – 351.
假设我们要求曲线在 x = 1 处的切线。先计算 y：y = (2×1+1)³(3×1-2)⁴ = (3)³(1)⁴ = 27。再用化简后的形式求 dy/dx：dy/dx = 42×1×(3)²×(1)³ = 42×9 = 378。切线斜率为 378，过点 (1, 27)。切线方程为：y – 27 = 378(x – 1)，即 y = 378x – 351。
This application highlights why a simplified derivative is advantageous – evaluating the derivative at a point becomes trivial.
这个应用凸显了化简导数的优势——在某点求导数值变得轻而易举。
9. Extending to Quotients | 拓展：商法则
If the exercise had been y = (2x+1)³ / (3x-2)⁴, we could rewrite it as y = (2x+1)³ (3x-2)⁻⁴ and use the product rule combined with the chain rule. Alternatively, apply the quotient rule directly: (u/v)’ = (u’v – uv’)/v². The same derivative techniques apply seamlessly.
若习题变为 y = (2x+1)³ / (3x-2)⁴，可改写为 y = (2x+1)³ (3x-2)⁻⁴，然后联用乘积法则与链式法则。也可直接运用商法则：(u/v)’ = (u’v – uv’)/v²。相同的求导技巧可无缝衔接。
Practising both approaches builds confidence and algebraic fluency, which is essential for IB Paper 2 and Paper 3.
两种方法都进行练习，可增强信心与代数流畅度，这对IB试卷二和试卷三至关重要。
10. Practice Variation | 变式练习
Try differentiating these similar functions to solidify your understanding:
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y = (x²+1)⁴ (2x-3)⁵
y = (x²+1)⁴ (2x-3)⁵
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y = (5x+2)² (4-3x)³
y = (5x+2)² (4-3x)³
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y = (1-2x)⁶ (x+3)⁷
y = (1-2x)⁶ (x+3)⁷
For each, apply the product rule first, then the chain rule for the individual powers, and simplify the result fully.
对每个函数，先运用乘积法则，再对各幂次使用链式法则，并将结果化简到底。
11. Connection to Optimization | 与优化问题的联系
Derivatives of such product functions often appear in optimization. Setting dy/dx = 0 gives critical points: 42x(2x+1)²(3x-2)³ = 0. Solutions: x = 0, x = -1/2, x = 2/3. By testing intervals or using the second derivative, one can classify these as local maxima, minima, or points of inflection.
此类乘积函数的导数常出现在优化问题中。令 dy/dx = 0 得到驻点：42x(2x+1)²(3x-2)³ = 0。解为 x = 0, x = -1/2, x = 2/3。通过区间测试或二阶导数，可将其分类为局部极大值、局部极小值或拐点。
For instance, at x = 0 the derivative changes from negative to positive, indicating a local minimum for the original function. This kind of analysis is central to IB exploration tasks.
例如，在 x = 0 处导数的符号由负变正，表明原函数有一个局部极小值。这种分析正是IB探究任务的核心。
12. Summary and Key Takeaways | 总结与关键要点
Exercise 22C.1 masterfully blends the product rule and chain rule. The key steps are: identify factors u and v, differentiate each using the chain rule, apply dy/dx = u’v + uv’, and then factor and simplify algebraically. Verification via logarithmic differentiation provides confidence. The resulting skill set transfers directly to harder problems involving trigonometric, exponential, or logarithmic composites.
习题22C.1精妙地融合了乘积法则与链式法则。关键步骤为：确定因式 u 和 v，用链式法则分别求导，应用 dy/dx = u’v + uv’，再进行代数分解与化简。用对数求导验证可增加把握。由此建立的技能可直接迁移到涉及三角函数、指数或对数复合的更复杂问题中。
Always check for common mistakes, practice with variations, and interpret the derivative in context – such as tangent slopes and optimization. Mastery of these fundamentals will serve you well across the IB Mathematics curriculum.
务必检查常见错误，用变式题目进行练习，并在实际问题情境中理解导数的意义——例如切线斜率和优化。熟练掌握这些基础，将让你在整个IB数学课程中受益匪浅。
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