The derivative · 导数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| derivative/dɪˈrɪvətɪv/ | 导数 | dǎo shù |
| gradient of the tangent/ˈɡreɪdɪənt ɒvðə ˈtændʒənt/ | 切线斜率 | qiè xiàn xié lǜ |
| differentiation from first principles/ˌdɪfəˌrenʃɪˈeɪʃn frɒm fɜːst ˈprɪnsɪplz/ | 从定义求导 | cóng dìng yì qiú dǎo |
| power rule/ˈpaʊə ruːl/ | 幂法则 | mì fǎ zé |
| product rule/ˈprɒdʌkt ruːl/ | 乘积法则 | chéng jī fǎ zé |
| quotient rule/ˈkwəʊʃənt ruːl/ | 商法则 | shāng fǎ zé |
| chain rule/tʃeɪn ruːl/ | 链式法则 | liàn shì fǎ zé |
| stationary point/ˈsteɪʃənəri pɔɪnt/ | 驻点 | zhù diǎn |
| second derivative/ˈsekənd dɪˈrɪvətɪv/ | 二阶导数 | èr jiē dǎo shù |
| maximum/ˈmæksɪməm/ | 极大值 | jí dà zhí |
| minimum/ˈmɪnɪməm/ | 极小值 | jí xiǎo zhí |
A finite average hides local change
- For $y=x^2$, the secant from x equal to 1 to 2 has slope 3, but the tangent at x equal to 1 has slope 2.
- A derivative 导数 is an instantaneous limiting rate, when it exists. Geometrically it gives the gradient of the tangent 切线斜率 at a differentiable point.
有限平均值掩盖局部变化
- 对于 $y=x^2$，从 x=1 到 x=2 的割线斜率为3，但在 x=1 处的切线斜率为2。
- 导数 是瞬时极限比率（若存在）。几何上，它给出可微点处的切线斜率。
Slide the second point towards the first · 将第二点向第一点滑动
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
The chord becomes the tangent, and its gradient becomes the derivative. · 当弦变为切线时，其斜率即变为导数。
What does the derivative of a function at a point measure? · 函数在某一点的导数衡量的是什么？
Where it exists, the derivative is an instantaneous limiting rate of change and tangent gradient. A definite integral measures signed accumulation, which need not equal geometric area. · 若存在导数，则其为瞬时极限变化率和切线斜率。定积分衡量有符号累积量，不一定等于几何面积。
Take a limit without dividing by zero
- Differentiation from first principles 从定义求导 uses $f'(x)=\lim_{h\to0}[f(x+h)-f(x)]/h$. The difference quotient uses nonzero h before its limit is taken.
- One-sided limits must agree for a two-sided derivative. A sharp corner can prevent differentiability even if the function is continuous.
取极限时避免除以零
- 从定义求导 使用 $f'(x)=\lim_{h\to0}[f(x+h)-f(x)]/h$。差商在取极限前需确保 h 不为零。
- 双侧导数存在的条件是左右极限必须一致。即使函数连续，尖角也可能导致不可微。
A power from its definition. For $f(x)=x^2$, the quotient is $[(x+h)^2-x^2]/h=2x+h$ for nonzero h. Its limit is $2x$. For $|x|$ at zero, the quotient $|h|/h$ instead has right-hand limit 1 and left-hand limit negative 1, so no derivative exists there.
基于定义的幂函数求导。 对于 $f(x)=x^2$，当 h 不为零时，差商为 $[(x+h)^2-x^2]/h=2x+h$，其极限为 $2x$。对于 $|x|$ 在零点处，差商 $|h|/h$ 的右极限为1，左极限为负1，因此该点不存在导数。
Select a rule for the function form
- The power rule 幂法则 gives $(x^n)'=nx^{n-1}$ on a valid domain. The product rule 乘积法则 gives $(uv)'=u'v+uv'$.
- The quotient rule 商法则 gives $(u/v)'=(u'v-uv')/v^2$ where v is nonzero. The chain rule 链式法则 gives $[f(g(x))]'=f'(g(x))g'(x)$, including the inner derivative.
根据函数形式选择法则
- 幂法则 在有效定义域内给出 $(x^n)'=nx^{n-1}$。乘积法则 给出 $(uv)'=u'v+uv'$。
- 商法则 在 v 不为零处给出 $(u/v)'=(u'v-uv')/v^2$。链式法则 给出 $[f(g(x))]'=f'(g(x))g'(x)$，包括内部函数的导数。
Differentiate y = x⁵. Give the derivative. · 对 y = x⁵ 求导。写出导数。
Bring the power down and reduce it by one: 5x⁴. · 将指数移下并减一：5x⁴。
Which rule differentiates y = (3x + 1)⁴? · 哪个法则用于对 y = (3x + 1)⁴ 求导？
A function inside a function. The derivative is 4(3x + 1)³ × 3 = 12(3x + 1)³. · 复合函数。导数为 4(3x + 1)³ × 3 = 12(3x + 1)³。
A stationary point needs classification
- A stationary point 驻点 has $f'(x)=0$. A nonzero second derivative 二阶导数 there gives a local maximum 极大值 when negative or minimum 极小值 when positive.
- A zero second derivative is inconclusive. Examine first-derivative signs or other evidence; $x^3$ has a stationary inflection at zero, not a maximum or minimum.
驻点需要分类判定
- 驻点 满足 $f'(x)=0$。若该点二阶导数不为零且为负，则为局部极大值；若为正，则为局部极小值。
- 二阶导数为零时无结论。需考察一阶导数符号或其他证据；$x^3$ 在零点处有驻点拐点，而非极大值或极小值。
y = x³ − 3x has a stationary point at a positive value of x. What is it? · y = x³ − 3x 在正 x 值处有一个驻点。它是什么？
dy/dx = 3x² − 3 = 0 gives x = ±1, so the positive one is x = 1. · dy/dx = 3x² − 3 = 0 解得 x = ±1，因此正值是 x = 1。
At a stationary point the second derivative is negative. What kind of point is it? · 若驻点处的二阶导数为负，这是什么类型的点？
Negative second derivative means the gradient is falling through zero, so the curve turns downward. · 负的二阶导数意味着斜率穿过零点下降，因此曲线向下弯曲。
Put the steps of a stationary-point question in the order that earns the marks. · 将驻点问题的步骤按得分顺序排列。
Find candidate x-values from the derivative, substitute into the original function for their coordinates, and justify classification. A zero second derivative is inconclusive. · 从导数中找到候选x值，代入原函数求坐标，并论证分类。二阶导数为零时结论不确定。
Find both coordinates and describe intervals. In sheet 3.1, factor the derivative to find candidate x-values, substitute into the original function for y, then test classification and where the curve increases or decreases.
求出两个坐标并描述区间。 在第3.1页中，对导数进行因式分解以找到候选x值，代入原函数求得y值，然后进行分类判定以及曲线递增或递减区间的测试。
$dy/dx$ differentiates with respect to x; a dot over y normally differentiates with respect to time. State the independent variable rather than treating these notations as interchangeable in every problem. Trigonometric derivative rules here use radians.
$dy/dx$ 是关于 x 的微分；y 上方的点通常表示对时间求导。应声明自变量，而不是在所有问题中将这两种记号视为可互换。此处三角函数求导规则使用弧度制。