📚 B9i Part 6: Applications of Differentiation | B9i 第六部分：微分的应用
This unit extends the basic rules of differentiation to solve problems involving tangents, normals, stationary points, optimisation, rates of change and kinematics. You will learn how to interpret the first and second derivatives and use them to analyse the shape and behaviour of functions.
本单元将微分的基本法则拓展到解决切线、法线、驻点、优化、变化率和运动学问题。你将学会如何解读一阶和二阶导数，并利用它们分析函数的形状和性质。
1. Review of Differentiation Rules | 微分法则回顾
Before applying differentiation, you must be confident with the core rules. For a power function, the derivative follows a simple pattern: multiply by the power and reduce the power by one.
在应用微分之前，你必须熟练掌握核心法则。对于幂函数，导数遵循一个简单规律：乘以指数，然后将指数减一。
If y = xⁿ, then dy/dx = n xⁿ⁻¹
The constant multiple rule states that the derivative of a f(x) is a f'(x). The sum rule states that you can differentiate term by term. For products and quotients, use the product and quotient rules. For composite functions, use the chain rule.
常数倍法则指出 a f(x) 的导数为 a f'(x)。和差法则指出你可以逐项求导。对于乘积和商，使用积法则和商法则。对于复合函数，使用链式法则。
- d/dx [3x⁴] = 12x³
- d/dx [5x² + 2x - 7] = 10x + 2
- d/dx [sin(2x)] = 2cos(2x)
These basic results are used in every applied differentiation problem, so memorise them accurately and practise recognising which rule applies.
这些基本结果会用于每一个微分应用问题，因此要准确记忆并练习识别该用哪条法则。
2. Tangents and Normals | 切线与法线
curve at a given point. If the curve is y = f(x), the tangent at x = a has gradient m = f'(a). The normal is perpendicular to the tangent, so its gradient is -1/m.
导数给出了曲线在某一点处切线的斜率。如果曲线为 y = f(x)，则在 x = a 处切线的斜率为 m = f'(a)。法线垂直于切线，因此其斜率为 -1/m。
Tangent gradient: m = dy/dx
Normal gradient: mnormal = -1 ÷ m
dx = 2x, so at x = 3 the gradient is 6. The tangent passes through (3, 9) with gradient 6:
例题：求 y = x² 在 x = 3 处的切线和法线。首先，y = 9 且 dy/dx = 2x，因此在 x = 3 处斜率为 6。切线经过点 (3, 9) 且斜率为 6：
y - 9 = 6(x - 3) ⇒ y = 6x - 9
- 3), which simplifies to y = -x/6 + 19/2. Always write the equation in the required form.
法线斜率为 -1/6，所以其方程为 y - 9 = -1/6 (x - 3)，化简为 y = -x/6 + 19/2。始终把方程写成题目要求的形式。
3. Increasing and Decreasing Functions | 函数的单调性
The sign of the first derivative tells you whether a function is increasing or decreasing on an interval. If dy/dx > 0, the function is increasing. If dy/dx < 0, the function is decreasing.
一阶导数的符号告诉你函数在某个区间上是递增还是递减。若 dy/dx > 0，则函数递增。若 dy/dx < 0，则函数递减。
dy/dx > 0 ⇒ increasing
dy/dx < 0 ⇒ decreasing
is increasing when x² - 1 > 0, which gives x < -1 or x > 1. It is decreasing when -1 < x < 1.
例如，y = x³ - 3x 的导数为 dy/dx = 3x² - 3 = 3(x² - 1)。当 x² - 1 > 0 即 x < -1 或 x > 1 时，函数递增。当 -1 < x < 1 时，函数递减。
This analysis helps you describe the behaviour of a curve and is essential for justifying stationary points in exam questions.
这一分析有助于你描述曲线的性质，也是在考试题中证明驻点性质所必需的。
4. Stationary Points | 驻点
tal. You must solve the equation dy/dx = 0 to find the x-coordinates of stationary points.
驻点出现在 dy/dx = 0 处。在该点切线是水平的。你必须解方程 dy/dx = 0 来求出驻点的 x 坐标。
Stationary points occur when dy/dx = 0
To classify a stationary point, you can use the first derivative test: examine the sign of dy/dx just before and just after the point. If the sign changes from positive to negative, it is a local maximum. If it changes from negative to positive, it is a local minimum. If the sign does not change, it is a point of inflection.
要对驻点进行分类，你可以使用一阶导数判别法：检查该点稍前和稍后 dy/dx 的符号。若符号由正变负，则为局部最大值。若由负变正，则为局部最小值。若符号不变，则为拐点。
maximum. At x = 1, dy/dx changes from negative to positive, so (1, -2) is a local minimum.
例题：y = x³ - 3x 的导数 dy/dx = 3x² - 3。令 dy/dx = 0 得 x = -1 和 x = 1。在 x = -1 处，dy/dx 由正变负，所以 (-1, 2) 是局部最大值。在 x = 1 处，dy/dx 由负变正，所以 (1, -2) 是局部最小值。
5. Second Derivative and Concavity | 二阶导数与凹凸性
The second derivative d²y/dx² measures the rate of change of the gradient. It tells you whether a curve is concave up or concave down. If d²y/dx² > 0, the curve is concave up. If d²y/dx² < 0, the curve is concave down.
二阶导数 d²y/dx² 度量斜率的变化率。它告诉你曲线是凹向上还是凹向下。若 d²y/dx² > 0，曲线凹向上。若 d²y/dx² < 0，曲线凹向下。
d²y/dx² > 0 ⇒ concave up
d²y/dx² < 0 ⇒ concave down
um. If d²y/dx² = 0, the test is inconclusive and you should use the first derivative test.
你还可以使用二阶导数对驻点进行分类。在驻点处，若 d²y/dx² > 0，则该点为局部最小值。若 d²y/dx² < 0，则该点为局部最大值。若 d²y/dx² = 0，该判别法无法确定，应使用一阶导数判别法。
. At x = 0, d²y/dx² = -6 < 0, so a maximum. At x = 2, d²y/dx² = 6 > 0, so a minimum.
例题：y = x³ - 3x² + 2 的导数 dy/dx = 3x² - 6x = 3x(x - 2)。驻点在 x = 0 和 x = 2。二阶导数为 d²y/dx² = 6x - 6。在 x = 0 处，d²y/dx² = -6 < 0，所以是最大值。在 x = 2 处，d²y/dx² = 6 > 0，所以是最小值。
6. Optimisation Problems | 优化问题
Optimisation problems ask you to find the maximum or minimum value of a quantity. You must write the quantity as a function of one variable, differentiate, set the derivative equal to zero, and verify that the answer is a maximum or minimum.
优化问题要求你求出某个量的最大值或最小值。你必须把该量写成一个变量的函数，对其求导，令导数为零，并验证答案是最值点。
Steps: model → differentiate → solve dy/dx = 0 → verify
ar sides be y. Then x + 2y = 20, so x = 20 - 2y. The area A = xy = y(20 - 2y) = 20y - 2y².
例题：一位农民靠着一堵直墙建一个矩形围栏，用 20 m 围栏围住另外三边。设与墙平行的边为 x，两条垂直边为 y。则 x + 2y = 20，所以 x = 20 - 2y。面积 A = xy = y(20 - 2y) = 20y - 2y²。
x = 10. Since d²A/dy² = -4 < 0, this is a maximum. The maximum area is 10 × 5 = 50 m².
求导得 dA/dy = 20 - 4y。令 dA/dy = 0 得 y = 5，然后 x = 10。由于 d²A/dy² = -4 < 0，这是最大值。最大面积为 10 × 5 = 50 m²。
7. Rates of Change | 变化率
he chain rule links their rates of change. This is often written as dA/dt = dA/dr × dr/dt.
许多现实中的量是相互关联的。如果两个量都随时间变化，链式法则将它们的变化率联系起来。这通常写作 dA/dt = dA/dr × dr/dt。
dA/dt = dA/dr × dr/dt
en dr/dt = 3. When r = 5 cm, the rate of change of area is dA/dt = 2π × 5 × 3 = 30π cm²/s.
例题：圆的面积为 A = πr²，所以 dA/dr = 2πr。若半径以 3 cm/s 的速度增加，则 dr/dt = 3。当 r = 5 cm 时，面积的变化率为 dA/dt = 2π × 5 × 3 = 30π cm²/s。
Always identify the known rate, the required rate, and the linking derivative before substituting values.
在代入数值之前，始终先确定已知速率、所求速率以及连接两者的导数。
8. Kinematics Applications | 运动学应用
In straight-line motion, differentiation connects displacement, velocity and acceleration. If displacement s is a function of time t, then velocity v is the first derivative and acceleration a is the second derivative.
在直线运动中，微分将位移、速度和加速度联系起来。若位移 s 是时间 t 的函数，则速度 v 是一阶导数，加速度 a 是二阶导数。
v = ds/dt and a = dv/dt = d²s/dt²
At t = 2 s, the velocity is v = 3(4) - 24 + 9 = -3 m/s and the acceleration is a = 0 m/s².
例题：一个质点运动，位移为 s(t) = t³ - 6t² + 9t 米，其中 t 以秒计。则 v(t) = 3t² - 12t + 9，a(t) = 6t - 12。在 t = 2 s 时，速度为 v = 3(4) - 24 + 9 = -3 m/s，加速度为 a = 0 m/s²。
t that instant. Use differentiation to find when the particle is at rest by solving v = 0.
速度为负意味着该瞬间质点沿负方向运动。通过解 v = 0 的方程，可以用微分求出质点何时静止。
9. Graph Sketching | 函数图像绘制
Differentiation is a powerful tool for sketching curves. You can find stationary points, intervals of increase or decrease, concavity, and intercepts. Combine these features to produce an accurate sketch.
微分是绘制函数图像的强大工具。你可以求出驻点、递增或递减区间、凹凸性以及截距。结合这些特征就能画出准确的草图。
Sketch checklist: intercepts → stationary points → concavity → shape
, the x-intercepts are found by solving x³ - 3x = 0, giving x(x² - 3) = 0, so x = 0, x = ±√3. The stationary points are (-1, 2) as a maximum and (1, -2) as a minimum. The curve rises, falls, then rises again.
对于 y = x³ - 3x，令 x³ - 3x = 0 求 x 截距，得 x(x² - 3) = 0，所以 x = 0，x = ±√3。驻点为 (-1, 2) 的最大值和 (1, -2) 的最小值。曲线先上升，再下降，然后又上升。
Sketching is often examined alongside stationary point calculations, so always label key points clearly.
图像绘制常常与驻点计算一起考查，因此务必清晰地标出关键点。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Students lose marks by forgetting to simplify before differentiating, misapplying the chain rule, or failing to verify whether a stationary point is a maximum or minimum. Always show clear working and state which derivative test you use.
学生常因求导前没有化简、误用链式法则、或未验证驻点是最大值还是最小值而失分。始终写出清晰的步骤，并说明你使用了哪个导数判别法。
- Simplify expressions before differentiating wherever possible.
- Use the chain rule for composite functions, not the product rule.
- Always substitute the x-value back into the original function to find the y-coordinate of a stationary point.
- Check the sign of the second derivative in optimisation problems to confirm maximum or minimum.
- Keep units in rates of change and kinematics answers.
在可能的情况下，求导前先化简表达式。对复合函数使用链式法则，而不是积法则。始终把 x 值代回原函数求出驻点的 y 坐标。在优化问题中检查二阶导数的符号以确认最大值或最小值。在变化率和运动学答案中保留单位。
Practice past-paper questions under timed conditions and mark your answers carefully using the official mark scheme.
在计时条件下练习历年真题，并使用官方评分标准仔细批改你的答案。
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