📚 Applications of More Complex Differentiation | 更复杂微分的应用
At A-level, differentiation moves beyond simple powers into composites, products, quotients, implicit curves and parametric motion. This article reviews the core rules and then applies them to the problem types most often tested in AQA Mathematics: nested functions, implicit relations, parametric curves, exponential models, connected rates of change and optimisation.
在 A-level 阶段，求导从简单幂函数扩展到复合函数、乘积、商、隐函数曲线和参数运动。本文回顾核心法则，并将其应用到 AQA 数学考试中最常考的题型：嵌套函数、隐函数关系、参数曲线、指数模型、相关变化率和最优化问题。
1. Review of the Core Rules | 核心法则回顾
For more complex differentiation you must be fluent with the chain rule, the product rule and the quotient rule. The chain rule is used whenever one function is inside another: differentiate the outside function, then multiply by the derivative of the inside function.
对于更复杂的求导，你必须熟练运用链式法则、乘积法则和商法则。只要一个函数嵌套在另一个函数内部，就要用到链式法则：先对外层函数求导，再乘以内层函数的导数。
d/dx [f(g(x))] = f′(g(x)) × g′(x)
The product rule handles u(x)v(x), and the quotient rule handles u/v. Notice the minus sign and the order of the terms in the quotient rule: the numerator must be v u′ − u v′.
乘积法则处理 u(x)v(x)，商法则处理 u/v。注意商法则中的负号与项的顺序：分子必须是 v u′ − u v′。
d/dx [u v] = u′v + u v′
d/dx [u/v] = (v u′ − u v′) / v²
You should also recall the derivatives of standard functions, because most complex questions combine them rather than using them alone.
你还应该记住标准函数的导数，因为大多数复杂题目是把它们组合起来，而不是单独使用。
| Function | 函数 | Derivative | 导数 |
|---|---|
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
2. The Chain Rule with Nested Functions | 嵌套函数的链式法则
on is the fifth power and the inside function is 3x² + 1, so the derivative is 5(3x² + 1)⁴ multiplied by 6x.
最常见的应用是线性或二次表达式的幂。对于 y = (3x² + 1)⁵，外层函数是五次幂，内层函数是 3x² + 1，因此导数是 5(3x² + 1)⁴ 乘以 6x。
dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴
), the derivative is 2/(2x + 1). For y = exp(sin x), the derivative is cos x × exp(sin x).
当内层函数是对数或指数时，结构相同。对于 y = ln(2x + 1)，导数为 2/(2x + 1)。对于 y = exp(sin x)，导数为 cos x × exp(sin x)。
A frequent exam error is to differentiate the outside function but forget to multiply by the derivative of the inside function. Always ask yourself: what is the inside function, and what is its derivative?
常见的考试错误是对外层函数求导后忘记乘以内部函数的导数。始终问自己：内层函数是什么，它的导数是什么？
3. Combining Product and Quotient Rules | 乘积与商法则的组合
he numerator or denominator must be differentiated using the chain rule. Take y = x² ln x.
许多 AQA 题目要求在乘积法则中有一个因子本身是复合函数，或在商法则中分子或分母需要用链式法则求导。以 y = x² ln x 为例。
dy/dx = 2x × ln x + x² × 1/x = 2x ln x + x = x(2 ln x + 1)
ves a numerator of x cos x − sin x over x². Check carefully that the term with the minus sign is the product of the original numerator and the derivative of the denominator.
对于 y = (sin x)/x，商法则给出的分子是 x cos x − sin x，分母是 x²。仔细检查带负号的项是否是原分子与分母导数的乘积。
dy/dx = (x cos x − sin x) / x²
A useful strategy is to simplify before differentiating only when it genuinely reduces the work. For example, (x³ + x)/x simplifies to x² + 1, which is far quicker than applying the quotient rule.
一个有用的策略是：只有在真正能减少计算量时才先化简再求导。例如，(x³ + x)/x 可化简为 x² + 1，这比使用商法则快得多。
4. Implicit Differentiation | 隐函数微分
When y is not given explicitly as a function of x, differentiate both sides with respect to x and remember that every derivative of a term in y must be multiplied by dy/dx. This is the chain rule applied to y as a function of x.
当 y 没有明确表示为 x 的函数时，对等式两边关于 x 求导，并记住每一项含 y 的导数都必须乘以 dy/dx。这就是把 y 看作 x 的函数后应用链式法则。
y term gives 2x + 2y dy/dx = 0, so dy/dx = −x/y. The gradient at (3, 4) is therefore −3/4.
考虑圆 x² + y² = 25。逐项求导得到 2x + 2y dy/dx = 0，因此 dy/dx = −x/y。在点 (3, 4) 处的梯度因此为 −3/4。
2x + 2y dy/dx = 0 → dy/dx = −x/y
+ 6x dy/dx. Collect the dy/dx terms on one side and factorise. Exam questions often then ask for the equation of the tangent or normal at a given point.
对于 x³ + y³ = 6xy 这样的曲线，逐项求导：3x² + 3y² dy/dx = 6y + 6x dy/dx。把含 dy/dx 的项移到一边并提取公因式。考题通常接着要求在给定点处求切线或法线方程。
5. Parametric Differentiation | 参数方程求导
the two derivatives with respect to t. This is often written as dy/dx = (dy/dt) ÷ (dx/dt).
当 x 和 y 都用参数 t 表示时，链式法则给出 dy/dx 等于两个关于 t 的导数之比。通常写作 dy/dx = (dy/dt) ÷ (dx/dt)。
dy/dx = (dy/dt) ÷ (dx/dt)
on of t, and the tangent gradient at a particular point is found by substituting the relevant value of t.
对于 x = t² 和 y = t³，dy/dt = 3t²，dx/dt = 2t，所以 dy/dx = 3t/2。这是 t 的函数，某一点处的切线梯度可通过代入相应的 t 值得到。
For a second derivative, differentiate dy/dx with respect to t and then divide by dx/dt again. This extra division is essential and is frequently forgotten in exams.
求二阶导数时，先对 dy/dx 关于 t 求导，再除以 dx/dt。这个额外的除法很关键，考试中经常被遗忘。
d²y/dx² = (d/dt [dy/dx]) ÷ (dx/dt)
6. Tangents, Normals and Stationary Points | 切线、法线与驻点
Differentiation is the tool for finding gradients. The tangent at a point has gradient dy/dx evaluated at that point, and the normal is perpendicular to the tangent, so its gradient is the negative reciprocal.
求导是求梯度的工具。某点处的切线梯度等于该点处 dy/dx 的值，而法线与切线垂直，因此其梯度为负倒数。
mnormal = −1 ÷ mtangent
6x = 0, giving x = 0 and x = 2. Substituting back gives the coordinates (0, 4) and (2, 0).
对于曲线 y = x³ − 3x² + 4，dy/dx = 3x² − 6x。驻点满足 3x² − 6x = 0，得到 x = 0 和 x = 2。代回原式得到坐标 (0, 4) 和 (2, 0)。
To classify a stationary point, use the second derivative: if d²y/dx² is positive the point is a minimum, if negative it is a maximum, and if zero further investigation is needed. A sign table for dy/dx around the point is the safest method when the second derivative vanishes.
判断驻点的类型可用二阶导数：若 d²y/dx² 为正，该点为极小值；若为负，则为极大值；若为零，则需要进一步研究。当二阶导数为零时，考察驻点两侧 dy/dx 的符号表是最稳妥的方法。
7. Exponential Growth and Decay Models | 指数增长与衰减模型
special property that their derivative is proportional to the function itself. For y = Aekt, differentiating gives dy/dt = kAekt = ky, where k is the rate constant.
指数函数具有特殊性质：其导数与函数本身成正比。对于 y = Aekt，求导得到 dy/dt = kAekt = ky，其中 k 是速率常数。
y = A ekt → dy/dt = k y
If k is positive the quantity grows, as in population models or compound interest. If k is negative the quantity decays, as in radioactive decay or the concentration of a drug in the bloodstream.
若 k 为正，则该量增长，例如人口模型或复利。若 k 为负，则该量衰减，例如放射性衰变或血液中药物浓度的下降。
Exam questions usually give two data points so that you can find A and k, then ask for a value at a future time or the time taken to halve or double. Always keep full accuracy until the final answer and state the units of time clearly.
考题通常给出两个数据点，让你求出 A 和 k，然后问未来某时刻的值，或减半、加倍所需的时间。始终保留完整精度直到最后一步，并清楚写出时间的单位。
8. Logarithmic Differentiation and General Powers | 对数求导与一般幂函数
When the variable appears in both the base and the exponent, take natural logarithms of both sides before differentiating. This converts powers into products and makes the chain rule straightforward.
当变量同时出现在底数和指数中时，先对两边取自然对数再求导。这样可把幂转化为乘积，使链式法则更易使用。
For y = xx gives ln y = x ln x. Differentiating implicitly gives (1/y) dy/dx = ln x + 1, so dy/dx = xx (ln x + 1).
对于 y = xx，取对数得 ln y = x ln x。隐式求导得 (1/y) dy/dx = ln x + 1，因此 dy/dx = xx (ln x + 1)。
y = xx → dy/dx = xx (ln x + 1)
The same method proves the general result for a constant base a: d/dx [ax] = ax e the extra factor ln e equals 1, which is exactly why e is the natural base for calculus.
同样的方法可以证明常数底数 a 的一般结果：d/dx [ax] = ax ln a。特别地，当 a = e 时，额外因子 ln e 等于 1，这正是 e 成为微积分自然底数的原因。
9. Trigonometric Applications | 三角函数的应用
chain rule gives dy/dx = 2 sin x cos x, which by the double-angle identity equals sin 2x.
三角函数求导通常与链式法则或恒等式结合出现。对于 y = sin²x，链式法则给出 dy/dx = 2 sin x cos x，由二倍角恒等式等于 sin 2x。
d/dx [sin²x] = 2 sin x cos x = sin 2x
and applying the quotient rule gives sec²x. This result is worth remembering because it appears in gradient and rate-of-change problems.
对于 y = tan x，把 tan x 写成 sin x / cos x 并应用商法则可得 sec²x。这个结果值得记住，因为它常出现在梯度与变化率问题中。
s −k sin(kx). Watch the negative sign for cosine and remember the factor k from the chain rule.
题目也可能涉及 y = sin(kx)，其导数为 k cos(kx)；或 y = cos(kx)，其导数为 −k sin(kx)。注意余弦的负号，并记住链式法则带来的因子 k。
10. Connected Rates of Change | 相关变化率
In a related-rates problem, two or more quantities change with time and are linked by a geometric or algebraic formula. Differentiate the linking formula with respect to time and use the chain rule to introduce the rates you know and the rate you need.
在相关变化率问题中，两个或多个量随时间变化，并由几何或代数公式联系起来。对联系公式关于时间求导，并用链式法则引入已知变化率和待求变化率。
volume increases at 100 cm³ s⁻¹ when r = 5 cm, then dr/dt = 100 ÷ (4π × 25) = 1/π cm s⁻¹.
对于球体，V = (4/3)πr³，关于 t 求导得 dV/dt = 4πr² dr/dt。若当 r = 5 cm 时体积以 100 cm³ s⁻¹ 增加，则 dr/dt = 100 ÷ (4π × 25) = 1/π cm s⁻¹。
dV/dt = 4πr² × dr/dt
Always substitute the numerical values only after differentiating symbolically. Substituting too early often destroys the structure of the chain rule and leads to errors in the final units.
务必先进行符号求导，再代入数值。过早代入往往会破坏链式法则的结构，并导致最终单位出错。
11. Optimisation Problems | 最优化问题
ng the constraint to eliminate the other variable, then differentiate and solve dy/dx = 0.
最优化意味着在约束条件下求最大值或最小值。把待优化的量写成单个变量的函数，用约束条件消去另一个变量，然后求导并解 dy/dx = 0。
we have h = (600 − 2x²)/(4x). The volume is V = x²h, which becomes a function of x alone.
例如，一个密闭长方体盒子底面为边长 x 的正方形，高为 h，总表面积为 600 cm²。由 2x² + 4xh = 600 得 h = (600 − 2x²)/(4x)。体积 V = x²h，于是化为仅含 x 的函数。
V(x) = x² × (600 − 2x²)/(4x) = (600x − 2x³)/4 = 150x − ½x³
firming a maximum. Always justify the nature of the stationary point and check that the answer is physically sensible.
求导得 dV/dx = 150 − (3/2)x²。令其为零得 x² = 100，所以 x = 10。二阶导数为 −3x，在 x = 10 时为负，确认为最大值。始终说明驻点的性质，并检查答案在物理上是否合理。
12. Exam Technique and Common Errors | 考试技巧与常见错误
Complex differentiation questions reward clear structure. Write down the rule you are using, identify u and v or the inside function, and show the substitution before simplifying. This earns method marks even if an arithmetic slip occurs later.
复杂求导题奖励清晰的结构。写出你使用的法则，标出 u 和 v 或内层函数，在化简前先展示代入过程。即使之后出现算术失误，这也能拿到方法分。
Common errors include: forgetting the chain rule factor; writing u v′ − v u′ in the quotient rule; omitting dy/dx when differentiating a y-term implicitly; forgetting to divide by dx/dt when finding d²y/dx² for parametric curves; and losing a negative sign when differentiating cosine or finding a normal gradient.
常见错误包括：忘记链式法则因子；商法则中写成 u v′ − v u′；对含 y 的项隐式求导时漏掉 dy/dx；参数曲线求 d²y/dx² 时忘记除以 dx/dt；以及求余弦导数或法线梯度时丢掉负号。
Finally, keep answers exact unless the question asks for a decimal. Leave surds, logarithms and π in exact form, and round only at the very end if a numerical answer is required.
最后，除非题目要求小数，否则答案保持精确形式。把根式、对数和 π 保留为精确形式，只在最后需要数值答案时才进行四舍五入。
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