📚 B9i Part 2: Applications of Differentiation | B9i 第二部分：微分法的应用
This article continues the B9i differentiation topic and focuses on applying the derivative to coordinate geometry, optimization and connected rates of change. You will revise standard rules, work through implicit and parametric differentiation, and learn how to justify maximum or minimum values clearly in an exam.
本文继续 B9i 微分法主题，重点讲解导数在坐标几何、最优化以及相关变化率中的应用。你将复习标准求导法则，练习隐函数与参数方程求导，并学会在考试中清晰地说明最大值或最小值的判定依据。
1. Review of Differentiation Rules | 求导法则回顾
Before tackling applications, make sure you can differentiate powers, exponentials, logarithms and trigonometric functions fluently. The derivative gives the gradient of a curve at any point, so it is the essential tool for tangents, normals, stationary points and optimization.
在进入应用之前，请确保你能够熟练地对幂函数、指数函数、对数函数和三角函数求导。导数给出曲线在任意一点的斜率，因此它是解决切线、法线、驻点和最优化问题的核心工具。
| f(x) | f'(x) |
|---|---|
| xⁿ | nxⁿ⁻¹ |
| eᵏˣ | keᵏˣ |
| ln x | 1/x |
| sin kx | k cos kx |
| cos kx | −k sin kx |
| tan kx | k sec² kx |
Keep these results on a formula card and test yourself under timed conditions. In many application questions, the first mark comes from writing down the derivative correctly, so fluency here gains easy marks.
把这些结果写在公式卡上，并在限时条件下自测。在许多应用题中，第一分往往来自正确写出导数，因此熟练记忆这些结果可以轻松得分。
2. The Chain Rule in Context | 链式法则在实际中的应用
n rule inside a power or a composite function. If y = [u(x)]ⁿ, then dy/dx = n[u(x)]ⁿ⁻¹ × u'(x). Do not forget the inner derivative.
许多考题把链式法则隐藏在幂函数或复合函数中。如果 y = [u(x)]ⁿ，则 dy/dx = n[u(x)]ⁿ⁻¹ × u'(x)。千万不要忘记内层导数。
if y = (3x² + 1)⁵, let u = 3x² + 1. Then du/dx = 6x, so dy/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴.
例如，如果 y = (3x² + 1)⁵，令 u = 3x² + 1。则 du/dx = 6x，因此 dy/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴。
dy/dx = 30x(3x² + 1)⁴
e chain rule simply multiplies by the coefficient of x. The result is dy/dx = 14(2x − 3)⁶.
当内层函数为线性函数时，例如 y = (2x − 3)⁷，链式法则只需乘以 x 的系数。结果是 dy/dx = 14(2x − 3)⁶。
3. Implicit Differentiation | 隐函数求导
When y is not given explicitly as a function of x, differentiate both sides of the equation with respect to x. Each time you differentiate a term involving y, multiply by dy/dx.
当 y 没有显式地表示为 x 的函数时，需要对等式两边同时关于 x 求导。每当你对包含 y 的项求导时，都要乘以 dy/dx。
= 25, differentiating both sides gives 2x + 2y dy/dx = 0. Rearranging gives dy/dx = −x/y.
对于圆 x² + y² = 25，两边求导得到 2x + 2y dy/dx = 0。整理后得到 dy/dx = −x/y。
x² + y² = 25 → 2x + 2y dy/dx = 0 → dy/dx = −x/y
Questions often ask for the gradient at a particular point. Substitute the x and y coordinates only after you have made dy/dx the subject. This avoids losing track of which variables are still present.
题目经常要求在某一点处的斜率。应先将 dy/dx 写成主项，再代入 x 和 y 的坐标。这样可以避免混淆哪些变量仍然存在。
4. Parametric Differentiation | 参数方程求导
= (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0. This result follows directly from the chain rule.
如果 x = f(t) 且 y = g(t)，则 dy/dx = (dy/dt) ÷ (dx/dt)，前提是 dx/dt ≠ 0。这一结果直接来自链式法则。
, given x = t² and y = 3t + 1, we have dx/dt = 2t and dy/dt = 3. Therefore dy/dx = 3/(2t).
例如，给定 x = t² 和 y = 3t + 1，我们有 dx/dt = 2t 和 dy/dt = 3。因此 dy/dx = 3/(2t)。
dy/dx = (dy/dt) ÷ (dx/dt)
parametric form. Use d²y/dx² = d/dx(dy/dx) = d/dt(dy/dx) ÷ dx/dt, again by the chain rule.
你可能还需要参数形式的二阶导数。使用 d²y/dx² = d/dx(dy/dx) = d/dt(dy/dx) ÷ dx/dt，这同样来自链式法则。
5. Second Derivatives | 二阶导数
The second derivative d²y/dx² tells you how the gradient itself is changing. It is central to determining the nature of a stationary point and to understanding concavity.
二阶导数 d²y/dx² 描述的是斜率本身的变化率。它是判断驻点性质以及理解曲线凹凸性的关键。
ond derivative is zero, so the sign table method may be needed to determine the nature of the stationary point.
如果 y = x³ − 6x² + 5x，则 dy/dx = 3x² − 12x + 5，且 d²y/dx² = 6x − 12。在 x = 2 处，二阶导数为零，因此可能需要使用符号表来判断该驻点的性质。
d²y/dx² = 6x − 12
In many exam solutions, the second derivative test is quicker: if f''(a) > 0, the point is a local minimum; if f''(a) < 0, it is a local maximum. If f''(a) = 0, the test is inconclusive.
在许多考试解法中，二阶导数检验更快：如果 f''(a) > 0，该点为局部极小值；如果 f''(a) < 0，该点为局部极大值。如果 f''(a) = 0，则检验无法判定。
6. Tangents and Normals | 切线与法线
The tangent to a curve at a point has gradient equal to the derivative at that point. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal −1/m.
曲线在一点处的切线斜率等于该点处的导数值。法线与切线垂直，因此其斜率为负倒数 −1/m。
nt equation is y − 5 = 4(x − 2). The normal has gradient −1/4, giving y − 5 = −1/4(x − 2).
对于 y = x² + 1 在 x = 2 处，斜率为 m = 4。该点为 (2, 5)，所以切线方程为 y − 5 = 4(x − 2)。法线斜率为 −1/4，方程为 y − 5 = −1/4(x − 2)。
Tangent: y − 5 = 4(x − 2) Normal: y − 5 = −1/4(x − 2)
Always state the negative reciprocal clearly. A common error is to use the same gradient for the normal as for the tangent, which loses method marks.
务必清楚地写出负倒数。一个常见错误是法线使用了与切线相同的斜率，这会失去方法分。
7. Stationary Points and Nature | 驻点及其判定
y point is a maximum, minimum or point of inflection, use either the second derivative or a sign table for dy/dx.
驻点出现在 dy/dx = 0 的位置。为了判断每个驻点是极大值、极小值还是拐点，可以使用二阶导数或 dy/dx 的符号表。
| Sign of dy/dx | Nature of stationary point |
|---|---|
| Positive → zero → negative | Local maximum |
| Negative → zero → positive | Local minimum |
| Same sign on both sides | Point of inflection |
For f'(x) = 0, calculate f''(x). If f''(x) > 0, the point is a minimum; if f''(x) < 0, it is a maximum. If f''(x) = 0, use the sign table instead.
当 f'(x) = 0 时，计算 f''(x)。如果 f''(x) > 0，该点为极小值；如果 f''(x) < 0，该点为极大值。如果 f''(x) = 0，则应改用符号表。
8. Optimization Problems | 最优化问题
Optimization questions ask you to maximize or minimize a quantity such as area, volume, profit or cost. The central step is to express the quantity as a function of one variable, then differentiate.
最优化问题要求你最大化或最小化某个量，例如面积、体积、利润或成本。核心步骤是把该量表示为单个变量的函数，然后求导。
- Assign letters to the variables in the question.
- Write a formula for the quantity to optimize.
- Use the given constraint to eliminate all but one variable.
- Differentiate and set the derivative equal to zero.
- Justify maximum or minimum and answer the specific question.
给题目中的变量设字母；写出需要优化的量的公式；利用给定约束消去除一个变量以外的所有变量；求导并令导数等于零；说明是最大值还是最小值，并回答具体问题。
A/dx = P/2 − 2x = 0 gives x = P/4. The maximum area occurs when the rectangle is a square.
对于周长为 P 的矩形，设长为 x，宽为 P/2 − x。面积为 A = x(P/2 − x) = (P/2)x − x²。则 dA/dx = P/2 − 2x = 0，解得 x = P/4。最大面积出现在矩形为正方形时。
dA/dx = P/2 − 2x = 0 → x = P/4
Many candidates find the x value but ignore the final justification. Use the second derivative or a sign test to confirm that the value gives a maximum, not a minimum.
许多考生求出了 x 值，却忽略了最后说明。应使用二阶导数或符号检验来确认该值对应最大值而非最小值。
9. Connected Rates of Change | 相关变化率
/dx × dx/dt. Identify which rate is given and which rate is required, then write the link.
使用链式法则将变化中的量联系起来：dy/dt = dy/dx × dx/dt。先确定已知的变化率和需要求的变化率，然后写出联系式。
h respect to time gives dA/dt = 2πr dr/dt. When r = 4 cm, dA/dt = 2π(4)(0.5) = 4π cm² s⁻¹.
对于一个圆形油膜，其半径以 dr/dt = 0.5 cm s⁻¹ 的速度增加，面积为 A = πr²。对时间求导得到 dA/dt = 2πr dr/dt。当 r = 4 cm 时，dA/dt = 2π(4)(0.5) = 4π cm² s⁻¹。
