Differentiation
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| differentiation/ˌdɪfəˌrenʃɪˈeɪʃn/ | Đạo hàm |
| power rule/ˈpaʊə ruːl/ | Quy tắc lũy thừa |
| chain rule/tʃeɪn ruːl/ | Quy tắc chuỗi |
| derivative/dɪˈrɪvətɪv/ | Hàm đạo hàm |
| stationary point/ˈsteɪʃənəri pɔɪnt/ | Điểm dừng |
| second derivative/ˈsekənd dɪˈrɪvətɪv/ | Đạo hàm bậc hai |
| optimisation/ˌɒptɪmaɪˈzeɪʃn/ | Tối ưu hóa |
The speed of change
- How fast is a rocket accelerating at the exact moment its fuel runs out? How steep is a roller coaster at its highest point?
- Differentiation 微分 answers these questions — it finds the instantaneous rate of change at any point on a curve.
The power rule 幂法则 and chain rule 链式法则
- Differentiation finds the gradient of a curve at each point — the derivative 导数 $f'(x)$ or $\dfrac{dy}{dx}$.
- The power rule: $\dfrac{d}{dx}(x^n) = n x^{n-1}$ (any rational $n$).
- Chain rule for a function inside a function: $\dfrac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$.

The derivative $\dfrac{dy}{dx} = 2x$ gives the gradient at any point: at $x = 2$, the gradient is $4$.
Worked example. $y = (3x+1)^4$. Let $u = 3x+1$, so $y = u^4$. Chain rule: $\dfrac{dy}{dx} = 4u^3 \cdot 3 = 12(3x+1)^3$.
The gradient at a point
gradient = dy/dx = f ′(x)
Slide the point along the curve — the tangent's slope IS the gradient f ′(x) there.
If y = x³, then dy/dx = 3x². What is the gradient at x = 2?
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dy/dx = 3x² = 3 × 2² = 3 × 4 = 12.
If y = (2x + 1)³, then dy/dx = 6(2x + 1)². What is dy/dx at x = 1?
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At x = 1: dy/dx = 6(2(1) + 1)² = 6(3)² = 6 × 9 = 54.
Stationary points 驻点
- A stationary point is where $\dfrac{dy}{dx} = 0$ (the tangent is flat).
- Test with the second derivative 二阶导数:
- $f''(x) > 0$ → minimum,
- $f''(x) < 0$ → maximum.
- Also: $\dfrac{dy}{dx} > 0$ → increasing; $< 0$ → decreasing.
Don't forget the chain rule. When differentiating $(ax+b)^n$, you must multiply by the derivative of the inside: $\dfrac{d}{dx}(ax+b)^n = an(ax+b)^{n-1}$, not just $n(ax+b)^{n-1}$.

At a maximum or minimum the gradient dy/dx is zero
The curve y = x² − 6x + 5 has a stationary point. At what x value?
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dy/dx = 2x − 6 = 0 gives x = 3.
At a stationary point, if the second derivative f″(x) > 0, the point is a:
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f″ > 0 means the curve bends upward — a minimum; f″ < 0 gives a maximum.
If dy/dx > 0 for all x in an interval, the function is increasing on that interval.
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A positive derivative means the function is rising — it is increasing.
For y = x³ − 6x² + 9x + 2, the second derivative at x = 3 is:
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d²y/dx² = 6x − 12. At x = 3: 6(3) − 12 = 18 − 12 = 6 > 0, so it's a minimum.
Worked example — finding stationary points
- $y = x^3 - 6x^2 + 9x + 2$.
- $\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x-1)(x-3)$.
- Stationary points at $x = 1$ and $x = 3$.
- $\dfrac{d^2y}{dx^2} = 6x - 12$: at $x=1$, $f'' = -6 < 0$ (maximum); at $x=3$, $f'' = 6 > 0$ (minimum).
Applications of differentiation
- Optimisation 最优化: find the maximum profit or minimum cost by setting $\dfrac{dy}{dx} = 0$.
- Rates of change: velocity is the derivative of displacement; acceleration is the derivative of velocity.
- Curve sketching: stationary points and the sign of $\dfrac{dy}{dx}$ reveal the shape.
- The derivative also gives the gradient of normals (perpendicular to the tangent).
You've got it
- power rule: $\dfrac{d}{dx}(x^n) = n x^{n-1}$; chain rule for nested functions
- stationary point: $\dfrac{dy}{dx} = 0$
- second-derivative test: $f'' > 0$ minimum, $f'' < 0$ maximum
- increasing when $\dfrac{dy}{dx} > 0$; decreasing when $\dfrac{dy}{dx} < 0$
Write from a minimum value of -16
- y = x cubed minus 12x.
- Find the gradient at x = 1 and the value at the positive stationary point.
For y = x cubed minus 12x, what is the gradient at x = 1, and what is y at the positive stationary point?
3 minus 12 is -9. At x = 2, 8 minus 24 is -16.
What is y at x = 2?
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The derivative is zero at x = 2 and the value is -16.