Chain, product, quotient and implicit differentiation · 链式法则、乘积法则、商法则和隐函数求导
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| chain rule/tʃeɪn ruːl/ | 链式法则 | liàn shì fǎ zé |
Does the inside expression change too?
- A cost curve is a power of a changing expression. Differentiating the outer power alone misses the rate of its input.
- This lesson studies chain rule 链式法则: The rule that multiplies the outer derivative by the inner derivative for a composite function.
内部表达式是否也在变化？
- 成本曲线是变化表达式的幂次形式。仅对外层幂次求导会遗漏其输入量的变化速率。
- 本课学习 chain rule 链式法则：针对复合函数，将外层函数的导数乘以内层函数的导数的规则。
Choose the mathematical structure
- For y=f(g(x)), y prime=f prime(g(x))g prime(x). For uv, differentiate to u prime v+uv prime. For u/v, use (u prime v-uv prime)/v², where v≠0.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 对于 y=f(g(x))，y prime=f prime(g(x))g prime(x)。对于 uv，求导得 u prime v+uv prime。对于 u/v，使用 (u prime v-uv prime)/v²，其中 v≠0。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines chain rule? · 以下哪项描述正确定义了链式法则？
The rule that multiplies the outer derivative by the inner derivative for a composite function. · 对于复合函数，将外层函数的导数乘以内层函数的导数的规则。
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For y=(3x+1)^4, y prime=4(3x+1)^3×3=12(3x+1)^3. At x=0, the gradient is 12. For x²+y²=25, differentiate implicitly: 2x+2y y prime=0, so y prime=-x/y when y≠0.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
对于 y=(3x+1)^4，y prime=4(3x+1)^3×3=12(3x+1)^3。在 x=0 处，斜率为 12。对于 x²+y²=25，隐函数求导得：2x+2y y prime=0，因此当 y≠0 时，y prime=-x/y。
Chain, product, quotient and implicit differentiation · 链式法则、乘积法则、商法则和隐函数求导
For y=f(g(x)), y prime=f prime(g(x))g prime(x) · 对于 y=f(g(x))，y'=f'(g(x))g'(x)
Compare the model with the worked case and explain one change. · 对比模型与已解案例，并说明其中一处变化。
Find the derivative of (3x+1)^4 at x=0. · 求 (3x+1)^4 在 x=0 处的导数。
Chain rule gives 4(3x+1)³×3; at zero this is 12. · 链式法则给出 4(3x+1)³×3；在零处该值为 12。
Test a tempting shortcut
- A derivative of a product is not the product of the derivatives. In implicit differentiation, every differentiated function of y brings a dy/dx factor. A quotient's denominator is squared.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The derivative of u(x)v(x) is always u prime times v prime. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 乘积的导数不等于导数的乘积。在隐函数求导中，每个关于 y 的求导项都会引入一个 dy/dx 因子。商的导数分母需平方。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
u(x)v(x) 的导数总是 u prime 乘以 v prime。此说法错误。请解释它违反了哪个定义或假设。
For x²+y²=25, find dy/dx at (3,4). · 对于 x²+y²=25，求在 (3,4) 处的 dy/dx。
Implicit differentiation gives 2x+2y y prime=0, so y prime=-3/4. · 隐函数求导得 2x+2y·y'=0，因此 y'=-3/4。
The derivative of u(x)v(x) is always u prime times v prime. · u(x)v(x) 的导数总是 u' 乘以 v'。
A derivative of a product is not the product of the derivatives. In implicit differentiation, every differentiated function of y brings a dy/dx factor. A quotient's denominator is squared. · 乘积的导数不等于导数的乘积。在隐函数求导中，每个对 y 的函数求导都会引入一个 dy/dx 因子。商法则的分母要平方。
Interpret a new situation
- Choose a useful form before differentiating: expanding a short polynomial may be simpler. For related rates, write the relation in symbols, differentiate with respect to time, then substitute measured values.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- 求导前选择一种便捷的形式：展开短多项式可能更简单。对于相关变化率问题，先用符号写出关系式，对时间求导，再代入测量数值。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
For y=xe^x, find dy/dx at x=0. · 对于 y=xe^x，求在 x=0 处的 dy/dx。
Product rule gives e^x+xe^x; at zero this is 1. · 乘积法则得 e^x+xe^x；在零处该值为 1。
Match each part of a complete solution to its purpose. · 将完整解答的每个部分与其目的相匹配。
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · 假设用于论证模型的合理性；检查用于验证结果；解释用于将其与问题建立联系。
Use this in your course
- edexcel IAL mathematics; official unit P3. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The rule that multiplies the outer derivative by the inner derivative for a composite function. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- Edexcel IAL 数学；官方单元 P3。其他单元的拓展内容已在范围审查中明确；不属于额外的加分要求。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
针对复合函数，将外层函数的导数乘以内层函数的导数的规则。选择正确的关系式，展示解题方法，验证其适用条件并解释结果。