📚 Application to More Complex Differentiation | 复杂微分法的应用
In AQA A-Level Mathematics, after mastering the basic rules of differentiation, you must apply them to more complex functions. This involves combining the product rule, quotient rule, and chain rule, often within a single expression, as well as handling implicit functions, parametric equations, and higher-order derivatives.
在 AQA A-Level 数学中，掌握基本微分法则之后，你需要将它们应用于更复杂的函数。这包括在同一条表达式中综合运用积法则、商法则和链式法则，以及处理隐函数、参数方程和高阶导数。
1. The Chain Rule — Repeated Use | 链式法则——重复使用
The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) × g'(x). In many complex problems, you must apply this rule more than once within a single differentiation. For example, to differentiate y = sin(cos x²), you first differentiate the outer sine, then the middle cosine, and finally the inner x².
链式法则指出：若 y = f(g(x))，则 dy/dx = f'(g(x)) × g'(x)。在许多复杂问题中，你需要在一次求导过程中多次运用该法则。例如，对 y = sin(cos x²) 求导时，首先对外层正弦求导，然后对中间层余弦求导，最后对内层 x² 求导。
When differentiating a power of a function, such as y = (3x² + 2x − 1)⁵, the rule simplifies to: multiply by the power, reduce the power by one, then multiply by the derivative of the inside function. The result is 5(3x² + 2x − 1)⁴ × (6x + 2), which can be factorised if required.
当对函数的幂求导时，例如 y = (3x² + 2x − 1)⁵，法则简化为：乘以幂次，幂次减一，再乘以内层函数的导数。结果为 5(3x² + 2x − 1)⁴ × (6x + 2)，如果需要，可以进一步因式分解。
d/dx [f(x)]ⁿ = n[f(x)]ⁿ⁻¹ × f'(x)
You should also recognise that trigonometric, exponential, and logarithmic functions each require their own chain-rule handling. For instance, d/dx [e^(sin x)] = e^(sin x) × cos x, and d/dx [ln(1 + x²)] = 2x/(1 + x²).
你还应该认识到，三角函数、指数函数和对数函数各自需要不同的链式法则处理。例如，d/dx [e^(sin x)] = e^(sin x) × cos x，而 d/dx [ln(1 + x²)] = 2x/(1 + x²)。
2. The Product Rule | 积法则
The product rule is used when differentiating a product of two functions of x. If y = uv, where u and v are both functions of x, then dy/dx = u(dv/dx) + v(du/dx). In AQA examinations, you are expected to apply this rule efficiently and simplify your final answer where appropriate.
当对两个关于 x 的函数的乘积求导时，使用积法则。若 y = uv，其中 u 和 v 都是 x 的函数，则 dy/dx = u(dv/dx) + v(du/dx)。在 AQA 考试中，你应当高效运用该法则，并在适当的情况下化简最终答案。
d/dx (uv) = u dv/dx + v du/dx
Consider y = x²·sin x. Here u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x. Applying the product rule gives dy/dx = x²·cos x + 2x·sin x = x(x cos x + 2 sin x). Taking out a common factor is often necessary to match the mark scheme.
考虑 y = x²·sin x。这里 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。运用积法则得 dy/dx = x²·cos x + 2x·sin x = x(x cos x + 2 sin x)。提取公因式往往是匹配评分标准所需的关键步骤。
For a product of three functions, y = uvw, the rule extends naturally: dy/dx = uv(dw/dx) + uw(dv/dx) + vw(du/dx). Although less common, you may encounter this in a multi-stage problem, so it is worth remembering.
对于三个函数的乘积，y = uvw，法则自然扩展为：dy/dx = uv(dw/dx) + uw(dv/dx) + vw(du/dx)。虽然这种情况不太常见，但你可能在多步骤问题中遇到，因此值得记住。
3. The Quotient Rule | 商法则
The quotient rule applies to functions expressed as a ratio of two differentiable functions. If y = u/v, then dy/dx = [v(du/dx) − u(dv/dx)] / v². This rule is essential for fractions such as y = tan x, which is derived as sin x / cos x.
商法则适用于表示为两个可微函数之比的函数。若 y = u/v，则 dy/dx = [v(du/dx) − u(dv/dx)] / v²。该法则对于处理诸如 y = tan x（即 sin x / cos x）这样的分式至关重要。
d/dx (u/v) = (v du/dx − u dv/dx) / v²
A common error to avoid is mixing up the order of subtraction. The numerator must be ‘v times du/dx minus u times dv/dx’. A memory aid is to remember the formula as ‘low d high minus high d low, over the square of the denominator’ — but in algebraic form: [v(u’) − u(v’)] / v².
需要避免的一个常见错误是混淆减法顺序。分子必须是”v 乘以 du/dx 减去 u 乘以 dv/dx”。可以借助口诀记忆，但在代数形式上应当准确写出：[v(u’) − u(v’)] / v²。
For example, differentiate y = (2x + 1) / (x² + 3). Let u = 2x + 1, du/dx = 2, and v = x² + 3, dv/dx = 2x. Then dy/dx = [(x² + 3)(2) − (2x + 1)(2x)] / (x² + 3)². Expanding the numerator gives 2x² + 6 − (4x² + 2x) = −2x² − 2x + 6, so dy/dx = (−2x² − 2x + 6) / (x² + 3)².
例如，对 y = (2x + 1) / (x² + 3) 求导。令 u = 2x + 1，du/dx = 2；v = x² + 3，dv/dx = 2x。则 dy/dx = [(x² + 3)(2) − (2x + 1)(2x)] / (x² + 3)²。展开分子得 2x² + 6 − (4x² + 2x) = −2x² − 2x + 6，因此 dy/dx = (−2x² − 2x + 6) / (x² + 3)²。
4. Combining Rules in One Expression | 在同一表达式中综合运用法则
AQA examination questions frequently require you to combine the product rule with the chain rule, or the quotient rule with the chain rule. For example, differentiating y = x³·e^(2x) requires both the product and chain rules simultaneously.
AQA 考试题经常要求你同时运用积法则和链式法则，或商法则与链式法则相结合。例如，对 y = x³·e^(2x) 求导需要同时使用积法则和链式法则。
Let u = x³ and v = e^(2x). Then du/dx = 3x² and dv/dx = 2e^(2x) (via the chain rule). Applying the product rule: dy/dx = x³·2e^(2x) + e^(2x)·3x² = x²·e^(2x)(2x + 3).
设 u = x³，v = e^(2x)。则 du/dx = 3x²，dv/dx = 2e^(2x)（通过链式法则）。运用积法则：dy/dx = x³·2e^(2x) + e^(2x)·3x² = x²·e^(2x)(2x + 3)。
Another typical combined example is y = (x² + 1)⁴·(2x − 1)³. Both factors require the chain rule before applying the product rule. Differentiating gives 4(x² + 1)³(2x)(2x − 1)³ + 3(2x − 1)²(2)(x² + 1)⁴. Factorising leads to (x² + 1)³(2x − 1)² [8x(2x − 1) + 6(x² + 1)].
另一个典型的综合例子是 y = (x² + 1)⁴·(2x − 1)³。两个因子在应用积法则前都需要使用链式法则。求导得 4(x² + 1)³(2x)(2x − 1)³ + 3(2x − 1)²(2)(x² + 1)⁴。因式分解后得到 (x² + 1)³(2x − 1)² [8x(2x − 1) + 6(x² + 1)]。
5. Implicit Differentiation | 隐函数求导
When y is not explicitly expressed in terms of x, we use implicit differentiation. Every term involving y is differentiated with respect to x, and each such term is multiplied by dy/dx. For example, differentiating y³ with respect to x gives 3y²(dy/dx).
当 y 未明确表示为 x 的函数时，我们使用隐函数求导。每一项含有 y 的项都关于 x 求导，并且每个这样的项都要乘以 dy/dx。例如，对 y³ 关于 x 求导得到 3y²(dy/dx)。
d/dx (yⁿ) = n·yⁿ⁻¹·(dy/dx)
Consider the equation x² + y² = 25. Differentiating both sides with respect to x yields 2x + 2y(dy/dx) = 0. Solving for dy/dx gives dy/dx = −x/y, which describes the gradient of the circle at any point (x, y).
考虑方程 x² + y² = 25。两边关于 x 求导得到 2x + 2y(dy/dx) = 0。解出 dy/dx 得 dy/dx = −x/y，这描述了圆上任意一点 (x, y) 处的斜率。
For a more complex implicit equation such as x³ + y³ = 6xy, differentiation gives 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). Rearranging collects all dy/dx terms on one side: dy/dx(3y² − 6x) = 6y − 3x², so dy/dx = (6y − 3x²) / (3y² − 6x).
对于更复杂的隐式方程，如 x³ + y³ = 6xy，求导得 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx)。整理后将所有含 dy/dx 的项移到一侧：dy/dx(3y² − 6x) = 6y − 3x²，因此 dy/dx = (6y − 3x²) / (3y² − 6x)。
6. Parametric Differentiation | 参数方程求导
When x and y are both given as functions of a parameter t, we differentiate each with respect to t and then use the relationship dy/dx = (dy/dt) / (dx/dt). This is a direct application of the chain rule and is a standard AQA Pure Mathematics topic.
当 x 和 y 都表示为参数 t 的函数时，我们分别对 t 求导，然后利用关系 dy/dx = (dy/dt) / (dx/dt)。这是链式法则的直接应用，也是 AQA 纯数学部分的标准考点。
dy/dx = (dy/dt) ÷ (dx/dt)
Suppose x = t² + 1 and y = t³ − t. Then dx/dt = 2t and dy/dt = 3t² − 1. Therefore dy/dx = (3t² − 1) / (2t), provided t ≠ 0. This result gives the gradient of the curve at any parameter value t.
假设 x = t² + 1，y = t³ − t。则 dx/dt = 2t，dy/dt = 3t² − 1。因此 dy/dx = (3t² − 1) / (2t)，前提是 t ≠ 0。该结果给出曲线上任意参数值 t 处的斜率。
In more demanding questions, you may need to find the equation of the tangent or normal at a specific parameter value. For this, substitute the given t into x, y, and dy/dx, then use the straight-line equation y − y₁ = m(x − x₁).
在更高要求的问题中，你可能需要求在特定参数值处的切线或法线方程。为此，将给定的 t 代入 x、y 和 dy/dx，然后使用直线方程 y − y₁ = m(x − x₁)。
7. Second Derivatives and Their Significance | 二阶导数及其意义
The second derivative, denoted d²y/dx², is obtained by differentiating dy/dx again with respect to x. It describes the rate of change of the gradient and determines the nature of stationary points. In AQA questions, you may be asked to compute d²y/dx² for implicit or parametric functions, which requires careful application of the rules you have already learned.
二阶导数，记作 d²y/dx²，通过对 dy/dx 再次关于 x 求导得到。它描述斜率的变化率，并用于判定驻点的性质。在 AQA 考题中，你可能需要对隐函数或参数函数计算 d²y/dx²，这需要仔细运用你已学过的法则。
For parametric equations, you cannot simply differentiate dy/dx with respect to t; you must use d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt). For example, if dy/dx = t² and dx/dt = 2t, then d²y/dx² = 2t ÷ 2t = 1.
对于参数方程，你不能简单地对 dy/dx 关于 t 求导；必须使用 d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt)。例如，若 dy/dx = t² 且 dx/dt = 2t，则 d²y/dx² = 2t ÷ 2t = 1。
For implicit functions, after obtaining dy/dx in terms of x and y, differentiate again with respect to x. Every term containing y must be differentiated implicitly. This often results in d²y/dx² expressed in terms of x, y, and dy/dx.
对于隐函数，在获得用 x 和 y 表示的 dy/dx 之后，再次关于 x 求导。每个含有 y 的项都必须进行隐式求导。这通常会导致 d²y/dx² 用 x、y 和 dy/dx 表达。
8. Applying Differentiation to Rates of Change | 微分在变化率中的应用
Complex differentiation is frequently applied to real-world problems involving related rates. The chain rule allows us to connect different rates: if A depends on r, and r depends on time t, then dA/dt = dA/dr × dr/dt. This is a core application in AQA examination papers.
复杂微分经常应用于涉及相关变化率的实际问题。链式法则使我们能够连接不同的变化率：若 A 依赖于 r，而 r 依赖于时间 t，则 dA/dt = dA/dr × dr/dt。这是 AQA 试卷中的核心应用。
For example, a spherical balloon is inflated so that its radius increases at a constant rate of 0.5 cm/s. Since volume V = (4/3)πr³, we have dV/dr = 4πr². Therefore dV/dt = dV/dr × dr/dt = 4πr² × 0.5 = 2πr² cm³/s. At r = 3 cm, the volume increases at 18π cm³/s.
例如，一个球形气球充气时，其半径以恒定速率 0.5 cm/s 增大。由于体积 V = (4/3)πr³，我们有 dV/dr = 4πr²。因此 dV/dt = dV/dr × dr/dt = 4πr² × 0.5 = 2πr² cm³/s。当 r = 3 cm 时，体积以 18π cm³/s 的速度增大。
In such problems, always identify the known rate, the target rate, and the intermediate relationship between variables. Set up the chain-rule equation explicitly before substituting numerical values. This structured approach is essential for obtaining full marks.
在此类问题中，始终要确定已知变化率、目标变化率以及变量之间的中间关系。在代入数值之前，明确地建立链式法则方程。这种结构化的方法是获得满分的关键。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply