The derivative
| English | Chinese | Pinyin |
|---|---|---|
| derivative | 导数 | dǎo shù |
| gradient of the tangent | 切线斜率 | qiè xiàn xié lǜ |
| differentiation from first principles | 从定义求导 | cóng dìng yì qiú dǎo |
| power rule | 幂法则 | mì fǎ zé |
| product rule | 乘积法则 | chéng jī fǎ zé |
| quotient rule | 商法则 | shāng fǎ zé |
| chain rule | 链式法则 | liàn shì fǎ zé |
| stationary point | 驻点 | zhù diǎn |
| second derivative | 二阶导数 | èr jiē dǎo shù |
| maximum | 极大值 | jí dà zhí |
| minimum | 极小值 | jí xiǎo zhí |
How fast, at this exact instant?
- Average speed over an hour is easy. Speed right now is a harder question, because in an instant nothing moves.
- A derivative 导数 answers it: the rate of change at a single point.
- Geometrically it is the gradient of the tangent 切线斜率 — the slope of the curve where it touches.
Slide the second point towards the first
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
The chord becomes the tangent, and its gradient becomes the derivative.
What does the derivative of a function at a point measure?
It is an instantaneous rate. The area is what integration gives.
From first principles
- The fraction is the gradient of a line through two points on the curve.
- Slide the second point towards the first and the line becomes the tangent.
- Differentiation from first principles 从定义求导 is examinable, even though the rules do the work afterwards.
Differentiate $y = x^2$ from first principles.
As $h \to 0$ this becomes $2x$.
Every term cancels except the ones carrying $h$, and letting those vanish is the whole idea. The algebra is short; the limit is the point.
Differentiate y = x⁵. Give the derivative.
Bring the power down and reduce it by one: 5x⁴.
Which rule differentiates y = (3x + 1)⁴?
A function inside a function. The derivative is 4(3x + 1)³ × 3 = 12(3x + 1)³.
The four rules
- Power rule 幂法则: $y = x^n$ gives $\dfrac{dy}{dx} = nx^{\,n-1}$.
- Product rule 乘积法则: $(uv)' = u'v + uv'$.
- Quotient rule 商法则: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
- Chain rule 链式法则: $y = f(g(x))$ gives $f'(g(x)) \cdot g'(x)$.
y = x³ − 3x has a stationary point at a positive value of x. What is it?
dy/dx = 3x² − 3 = 0 gives x = ±1, so the positive one is x = 1.
At a stationary point the second derivative is negative. What kind of point is it?
Negative second derivative means the gradient is falling through zero, so the curve turns downward.
Put the steps of a stationary-point question in the order that earns the marks.
The classification is the marked step most often missed, and the question is not finished without it.
Stationary points
- A stationary point 驻点 is where $f'(x) = 0$: the curve is momentarily flat.
- The second derivative 二阶导数 classifies it. Negative means a maximum 极大值, positive a minimum 极小值.
- This pair of steps — solve $f'(x)=0$, then test $f''(x)$ — is the most examined procedure in the unit.
Setting the derivative to zero finds the point; the second derivative says what kind it is. Skipping the second step is the most common way to lose marks on a question you have essentially solved.
$\dfrac{dy}{dx}$, $f'(x)$ and $\dot{y}$ all mean the derivative, but a question that uses one expects it back. Notation is marked at this level, and switching mid-solution reads as not knowing they are the same thing.