Differentiation · 微分
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| calculus/ˈkælkjʊləs/ | 微积分 | wēi jī fēn |
| derivative/dɪˈrɪvətɪv/ | 导数 | dǎo shù |
| differentiate/ˌdɪfəˈrenʃɪeɪt/ | 求导 | qiú dǎo |
| stationary point/ˈsteɪʃənəri pɔɪnt/ | 驻点 | zhù diǎn |
The gradient that changes
- A straight line has one gradient everywhere. A curve has a different gradient at every point.
- Newton and Leibniz invented calculus 微积分 to answer: "What is the gradient of a curve right here?" The answer is the derivative 导数.
The power rule
- If $y = ax^n$, then $\dfrac{dy}{dx} = an\,x^{n-1}$.
- In words: bring the power down in front, then reduce it by one.
$y = x^3 + 2x^2 - 5x \Rightarrow \dfrac{dy}{dx} = 3x^2 + 4x - 5$. Differentiate 求导 term by term — each term is treated independently.
The gradient of a curve at a point is the gradient of the tangent there, which is what differentiation finds
The gradient at a point · 一点处的梯度
gradient = dy/dx
Slide the point — the tangent · 相切's slope is the gradient there. · 滑动这个点——切线的斜率是那里的斜率。
If y = x³ + 2x² − 5x, then dy/dx = 3x² + 4x − 5. What is its value at x = 1? · 如果 y = x³ + 2x² − 5x,那么 dy/dx = 3x² + 4x − 5。它在 x = 1 的值是多少?
3(1)² + 4(1) − 5 = 3 + 4 − 5 = 2.
For y = axⁿ, where n is a positive integer, differentiating gives: · 对于 y = axⁿ（其中 n 为正整数），求导得：
Bring the power down and reduce it by 1: a·n·xⁿ⁻¹. · 把幂拿下来并把它减 1:a·n·xⁿ⁻¹。
If y = 4x³, what is dy/dx at x = 2? · 如果 y = 4x³,dy/dx 在 x = 2 是多少?
dy/dx = 12x². At x = 2: 12(4) = 48. · dy/dx = 12x²。在 x = 2:12(4) = 48。
The derivative of y = x⁵ is dy/dx = ______. · y = x⁵ 的导数是 dy/dx = ______。
Power rule: bring down the 5, reduce the power: 5x⁴. · 幂法则:把 5 拿下来,减幂:5x⁴。
Special cases
- $y = c$ (a constant) → $\dfrac{dy}{dx} = 0$ (a horizontal line has zero gradient).
- $y = cx$ → $\dfrac{dy}{dx} = c$ (a straight line has constant gradient).
- The IGCSE power rule here uses nonnegative integer powers and sums of at most three terms; differentiating reciprocal functions is outside this syllabus.
Don't forget the constant term. The derivative of $x^2 + 7$ is $2x$, not $2x + 7$. The derivative of a constant is always zero.
The derivative of y = x² + 7 is 2x + 7. · y = x² + 7 的导数是 2x + 7。
The derivative of a constant is 0. So dy/dx = 2x, not 2x + 7. · 一个常数的导数是 0。所以 dy/dx = 2x,不是 2x + 7。
Stationary points 驻点
- A stationary point is where $\dfrac{dy}{dx} = 0$ — the curve is momentarily flat.
- $y = x^2 - 6x + 5 \Rightarrow \dfrac{dy}{dx} = 2x - 6 = 0 \Rightarrow x = 3$.
- At $x = 3$: $y = 9 - 18 + 5 = -4$. Turning point: $(3, -4)$.
A stationary point is where the curve is momentarily flat — the gradient (derivative) equals zero.
The curve y = x² − 6x + 5 has a turning point. At what x value (dy/dx = 0)? · 曲线 y = x² − 6x + 5 有一个转折点。在什么 x 值(dy/dx = 0)?
dy/dx = 2x − 6 = 0 gives x = 3. · dy/dx = 2x − 6 = 0 给出 x = 3。
Why differentiation matters
- Physics: velocity is the derivative of position; acceleration is the derivative of velocity.
- Economics: marginal cost is the derivative of total cost — the cost of one more unit.
- Optimisation: to maximise profit or minimise cost, set the derivative to zero.
A tangent estimates a curve gradient
- Draw a tangent touching the curve at the required point, then choose two clear points on the tangent, not two arbitrary points on the curve. Calculate its rise divided by run.
- For $y=x^2$ at $(2,4)$, a tangent through $(1,0)$ and $(3,8)$ has $m=(8-0)/(3-1)=4$. Differentiation confirms $dy/dx=2x=4$ at $x=2$.
A tangent passes through (1,0) and (3,8). Find its gradient. · 一条切线经过点 (1,0) 和 (3,8)。求其斜率。
Rise 8 divided by run 2 gives 4. · 上升量 8 除以水平距离 2 得到 4。
Minimum or maximum?
- For $y=x^2-6x+5$, the derivative is $2x-6$. It is negative at $x=2$, zero at 3 and positive at 4: falling then rising means a minimum at $(3,-4)$.
- For $y=-x^2+4x+1$, $dy/dx=-2x+4$. Positive at 1 and negative at 3 means a maximum at $x=2$, where $y=5$.
For y = −x² + 4x + 1, find the y-coordinate of its maximum. · 对于 y = −x² + 4x + 1，求其最大值对应的 y 坐标。
dy/dx = −2x + 4 is zero at x = 2; y = −4 + 8 + 1 = 5. · dy/dx = −2x + 4 在 x = 2 处为零；y = −4 + 8 + 1 = 5。
You've got it
- power rule: $y = ax^n \Rightarrow \dfrac{dy}{dx} = an\,x^{n-1}$
- differentiate a sum term by term; the derivative of a constant is zero
- a stationary point is where $\dfrac{dy}{dx} = 0$
- $y = x^3 + 2x^2 - 5x \Rightarrow \dfrac{dy}{dx} = 3x^2 + 4x - 5$