📚 Gradients, Tangents and Normals | 梯度、切线与法线
In A-Level Edexcel Mathematics, differentiation gives much more than a rate of change. It provides the gradient of a curve at any chosen point, and this gradient is the key to writing equations of tangents and normals. Whether the curve is given explicitly, implicitly, or parametrically, the same core ideas always apply. This article breaks down the methods step by step, with worked examples and common pitfalls.
在 A-Level Edexcel 数学中，微分给出的不仅仅是变化率。它提供了曲线上任意选定点的梯度，而这个梯度是写出切线和法线方程的关键。无论曲线是以显式、隐式还是参数形式给出，相同的核心思想始终适用。本文将逐步分解这些方法，并配有例题和常见错误提醒。
1. The derivative as a gradient function | 导数作为梯度函数
radient at a specific point (x₁, y₁), you substitute x = x₁ into the expression for dy/dx.
对于方程为 y = f(x) 的曲线，导数 dy/dx 是一个函数，它给出曲线在每个 x 值处的梯度。从几何上看，dy/dx 表示曲线在某一点处切线的斜率。要找到特定点 (x₁, y₁) 处的梯度，你需要将 x = x₁ 代入 dy/dx 的表达式中。
m = dy/dx at x = x₁
original curve, so y₁ is found by substituting x₁ into y = f(x), not into the derivative.
这个值 m 就是切线斜率。重要的是不要将斜率 m 与点本身混淆。点必须位于原曲线上，所以 y₁ 要通过将 x₁ 代入 y = f(x) 来求得，而不是代入导数。
2. Differentiation rules you need | 所需的微分法则
= n xⁿ⁻¹. You can differentiate term by term, so y = 4x³ − 2x + 5 gives dy/dx = 12x² − 2.
Edexcel 中关于切线和法线的题目要求你熟练进行微分。幂法则指出，如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹。你可以逐项求导，因此 y = 4x³ − 2x + 5 得到 dy/dx = 12x² − 2。
3x − 1)³. The product rule and quotient rule are needed for products and rational functions.
对于复合函数，链式法则是必不可少的。例如，如果 y = (3x − 1)⁴，那么 dy/dx = 4(3x − 1)³ × 3 = 12(3x − 1)³。对于乘积和有理函数，需要使用乘法法则和除法法则。
d/dx (uv) = u dv/dx + v du/dx d/dx (u/v) = (v du/dx − u dv/dx) / v²
Being fluent in these rules saves time and reduces sign errors when you move on to finding tangent and normal equations.
熟练运用这些法则可以节省时间，并在接下来求切线和法线方程时减少符号错误。
3. Equation of a tangent | 切线方程
The tangent to a curve at a point is the straight line that touches the curve without crossing it at that point, and it has the same gradient as the curve there. If the point is (x₁, y₁) and the tangent gradient is m, then the equation of the tangent is:
曲线在某点的切线是与曲线在该点相接但不穿过曲线、并且在该点与曲线具有相同斜率的直线。如果点是 (x₁, y₁)，切线斜率为 m，那么切线方程为：
y − y₁ = m(x − x₁)
Always find y₁ by substituting x₁ into the original curve equation. A common mistake is to substitute x₁ into dy/dx and use that number as y₁. The derivative gives the gradient, not the y-coordinate of the point.
务必通过将 x₁ 代入原曲线方程来求 y₁。一个常见错误
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