Derivatives and stationary points · Extension · 导数与驻点 · 拓展
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| derivative/dɪˈrɪvətɪv/ | 导数 | dǎo shù |
What is the slope at one point?
- A curved road has different slopes at different positions. An average gradient cannot describe every point.
- This lesson studies derivative 导数: The instantaneous rate of change, also the gradient of a tangent.
某一点的斜率是什么？
- 弯曲道路在不同位置的斜率各不相同。平均梯度无法描述每一点的情况。
- 本课学习导数（derivative）：瞬时变化率，亦即切线的斜率。
Choose the mathematical structure
- For a polynomial term ax^n with nonnegative integer n, the gradient term is anx^(n-1). Add the differentiated terms. A stationary point has zero gradient.
- 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
- 对于多项式项 ax^n（其中 n 为非负整数），其导数项为 anx^(n-1)。将求导后的各项相加。驻点处的斜率为零。
- 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 derivative? · 以下哪项描述正确定义了导数？
The instantaneous rate of change, also the gradient of a tangent. · 瞬时变化率，亦即切线的斜率。
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=x³-3x, dy/dx=3x²-3. At x=1, the gradient is 0 and y=-2. The second derivative is 6x, positive at x=1, so this is a local minimum. At x=-1, y=2 and the second derivative is negative, giving a local maximum.
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=x³-3x，dy/dx=3x²-3。在 x=1 处，斜率为 0，y=-2。二阶导数为 6x，在 x=1 处为正，因此这是局部极小值。在 x=-1 处，y=2，二阶导数为负，给出局部极大值。
Derivatives and stationary points · 导数与驻点
For a polynomial term ax^n with nonnegative integer n, the gradient term is anx^(n-1) · 对于多项式项 ax^n（其中 n 为非负整数），其梯度项为 anx^(n-1)
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。
For y=x³-3x, find dy/dx at x=2. · 对于 y=x³-3x，求 x=2 处的 dy/dx。
dy/dx=3x²-3. At x=2 it is 12-3=9. · dy/dx=3x²-3。在 x=2 处，其值为 12-3=9。
Test a tempting shortcut
- A zero derivative does not always mean a maximum or minimum: y=x³ is stationary at 0 but continues increasing. An endpoint can also produce an extreme value on a restricted domain.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every point with zero derivative is a local maximum or minimum. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 导数为零并不总是意味着极大值或极小值：y=x³ 在 0 处是驻点，但函数继续递增。端点也可以在受限定义域上产生极值。
- 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。
所有导数为零的点都是局部极大值或极小值。此说法错误。请解释它违反了哪个定义或假设。
Find the y-coordinate of the stationary point at x=1. · 求 x=1 处驻点的 y 坐标。
Substitute into the original curve: y(1)=1-3=-2. · 代入原曲线方程：y(1)=1-3=-2。
Every point with zero derivative is a local maximum or minimum. · 导数为零的点不一定是局部最大值或最小值。
A zero derivative does not always mean a maximum or minimum: y=x³ is stationary at 0 but continues increasing. An endpoint can also produce an extreme value on a restricted domain. · 导数为零并不总是意味着最大值或最小值：y=x³ 在 0 处是驻点，但函数仍持续递增。端点在受限定义域上也可能产生极值。
Interpret a new situation
- This advanced-tier IGCSE lesson is limited to polynomial differentiation, tangent gradients and stationary points. Chain, product, quotient and implicit differentiation are excluded.
- 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
- 这门面向高级别的 IGCSE 课程仅涵盖多项式求导、切线斜率和驻点。不包括链式法则、乘积法则、商法则及隐函数求导。
- 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=4x³, find dy/dx at x=2. · 对于 y=4x³，求 x=2 处的 dy/dx。
The derivative is 12x²; at 2 it is 12×4=48. · 导数为 12x²；在 2 处，其值为 12×4=48。
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
- 9260 · Extension · 3.2. Match the target tier and specification before assigning extensions.
- 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 instantaneous rate of change, also the gradient of a tangent. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- 9260 · 拓展内容 · 3.2。在分配拓展内容前，请先匹配目标层级和课程大纲要求。
- 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。
瞬时变化率，即切线的斜率。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。