Differentiation: Rules, Techniques and Exam Applications | 微分：法则、技巧与考试应用

📚 Differentiation: Rules, Techniques and Exam Applications | 微分：法则、技巧与考试应用

Differentiation is one of the central topics in Edexcel A-Level Pure Mathematics. It allows us to find gradients, rates of change, maximum and minimum values, and the behaviour of curves. A strong command of differentiation rules and their applications is essential for both the pure and applied parts of the course.

微分是爱德思 A-Level 纯数学的核心主题之一。它帮助我们求梯度、变化率、最大值与最小值以及曲线的行为。扎实掌握微分法则及其应用，对纯数学和应用数学部分都至关重要。

1. Gradient and the Derivative | 梯度与导数

ction. For a curve y = f(x), the gradient of the tangent at a point is given by dy/dx or f'(x). This value tells us how fast y is changing relative to x at that exact point.

导数衡量函数在某一点的瞬时变化率。对于曲线 y = f(x)，切线的斜率由 dy/dx 或 f'(x) 给出。这个值告诉我们在该点处 y 相对于 x 的变化速度。

In Edexcel exam questions, you are often asked to interpret dy/dx as a gradient, find where a curve is increasing or decreasing, and use the derivative to describe the shape of a graph.

在爱德思考试题中，你经常需要将 dy/dx 解释为梯度，判断曲线在哪里递增或递减，并用导数描述图像的形状。

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2. Differentiation from First Principles | 从第一性原理求导

The formal definition of the derivative uses a limit. For a function f(x), its derivative is defined as the limit of the difference quotient as h approaches zero.

导数的正式定义使用极限。对于函数 f(x)，其导数定义为当 h 趋近于零时差商的极限。

f'(x) = lim (h→0) [f(x+h) − f(x)] / h

For example, if f(x) = x², then f'(x) = lim (h→0) [(x+h)² − x²] / h = lim (h→0) (2xh + h²) / h = lim (h→0) (2x + h) = 2x.

例如，若 f(x) = x²，则 f'(x) = lim (h→0) [(x+h)² − x²] / h = lim (h→0) (2xh + h²) / h = lim (h→0) (2x + h) = 2x。

Edexcel often tests this definition to check that you understand the limit process, not just the shortcut rules.

爱德思经常考查这一定义，以检验你是否理解极限过程，而不仅仅是套用速算法则。

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3. Basic Rules: Power, Constant and Sum | 基本法则：幂、常数与和

x = n xⁿ⁻¹. Constants differentiate to zero, and you can differentiate a sum term by term.

幂法则是最常用的求导工具。若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。常数求导为零，且和函数可以逐项求导。

If y = xⁿ, then dy/dx = n xⁿ⁻¹

is dy/dx = 12x³ − 4x. You should also know the derivatives of standard functions such as eˣ, ln x, sin x and cos x.

例如 y = 3x⁴ − 2x² + 5 的导数为 dy/dx = 12x³ − 4x。你还应掌握标准函数的导数，如 eˣ、ln x、sin x 和 cos x。

f(x)	f'(x)
xⁿ	n xⁿ⁻¹
eˣ	eˣ
ln x	1/x
sin x	cos x
cos x	−sin x

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4. Product Rule and Quotient Rule | 乘法法则与商法则

erentiating the product of two functions, use the product rule. If y = u v, then dy/dx = u' v + v' u. For example, if y = x² sin x, then dy/dx = 2x sin x + x² cos x.

当对两个函数的乘积求导时，使用乘法法则。若 y = u v，则 dy/dx = u' v + v' u。例如，若 y = x² sin x，则 dy/dx = 2x sin x + x² cos x。

Product rule: dy/dx = u' v + v' u

ng one function divided by another, use the quotient rule. If y = u / v, then dy/dx = (v u' − u v') / v². This rule is commonly tested with rational, logarithmic and exponential functions.

当对一个函数除以另一个函数求导时，使用商法则。若 y = u / v，则 dy/dx = (v u' − u v') / v²。该法则常在有理函数、对数函数和指数函数中考查。

Quotient rule: dy/dx = (v u' − u v') / v²

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5. Chain Rule and Implicit Differentiation | 链式法则与隐函数求导

The chain rule is used for composite functions. If y = f(g(x)), then dy/dx = f'(g(x)) g'(x). In Leibniz notation, dy/dx = dy/du × du/dx, where u = g(x).

链式法则用于复合函数。若 y = f(g(x))，则 dy/dx = f'(g(x)) g'(x)。用莱布尼茨记号表示为 dy/dx = dy/du × du/dx，其中 u = g(x)。

Chain rule: dy/dx = dy/du × du/dx

, let u = 2x + 1. Then dy/du = 5u⁴ and du/dx = 2, so dy/dx = 5(2x + 1)⁴ × 2 = 10(2x + 1)⁴.

例如，若 y = (2x + 1)⁵，设 u = 2x + 1。则 dy/du = 5u⁴，du/dx = 2，所以 dy/dx = 5(2x + 1)⁴ × 2 = 10(2x + 1)⁴。

spect to x, treating y as a function of x. This gives 2x + 2y(dy/dx) = 0, so dy/dx = −x/y.

对于 x² + y² = 25 等隐式方程，两边对 x 求导，并将 y 看作 x 的函数。可得 2x + 2y(dy/dx) = 0，因此 dy/dx = −x/y。

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6. Second Derivative and Stationary Points | 二阶导数与驻点

nima, or points of inflection. To classify them, you can use the second derivative d²y/dx².

驻点出现在 dy/dx = 0 处。它们可能是局部极大值、局部极小值或拐点。要判断它们的类型，可以使用二阶导数 d²y/dx²。

²y/dx² < 0, the point is a local maximum. If d²y/dx² = 0, further investigation is needed.

如果 dy/dx = 0 且 d²y/dx² > 0，该点为局部极小值。如果 dy/dx = 0 且 d²y/dx² < 0，该点为局部极大值。如果 d²y/dx² = 0，则需要进一步判断。

d²y/dx² > 0 ⇒ minimum; d²y/dx² < 0 ⇒ maximum

= x³ − 3x, then dy/dx = 3x² − 3 and d²y/dx² = 6x. Setting dy/dx = 0 gives x = 1 and x = −

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