📚 Mastering Tangents, Normals and Stationary Points (AQA Exercise 6H) | 掌握切线、法线与驻点（AQA 练习 6H）
Exercise 6H in the AQA A-level Mathematics differentiation chapter is a mixed exercise. It brings together gradient calculations, equations of tangents and normals, and the location of stationary points in one problem set, so the questions test whether you can move fluently between algebraic differentiation and its geometric interpretation on a curve.
AQA A-level 数学微分章节中的练习 6H 是一组综合练习。它把梯度计算、切线方程、法线方程以及驻点位置整合在同一套题目中，用来检验你能否在代数求导与曲线几何意义之间流畅转换。
The exercise assumes you are confident with the power rule, with rewriting roots and reciprocals as powers, and with the straight-line equation. Most marks are lost not through lack of knowledge but through small slips: a missed negative sign, a y-coordinate taken from the derivative instead of the original curve, or a normal gradient that has not been turned upside down.
本练习要求你熟练掌握幂函数求导法则、把根式和倒数改写成幂的形式，以及直线方程。丢分往往不是因为不会，而是因为小失误：漏掉负号、用导数而不是原函数求 y 坐标，或者忘记把切线梯度取负倒数来得到法线梯度。
1. What Exercise 6H Tests | 练习 6H 考查什么
Exercise 6H usually asks you to differentiate a given function f(x) or y e following: the gradient of the curve at a stated point, the equation of the tangent at that point, the equation of the normal at that point, or the coordinates and nature of the stationary points.
练习 6H 通常要求你先对给定函数 f(x) 或 y = ... 求导，然后求以下一项或多项：曲线在指定点处的梯度、该点处的切线方程、该点处的法线方程，或驻点的坐标与性质。
The reliable method has three steps. First, differentiate correctly to obtain dy/dx. Second, substitute the required x-value into dy/dx to obtain the numerical gradient m. Third, apply the straight-line equation y - y₁ = m(x - x₁), having found y₁ by substituting the x-value into the original equation, never into the derivative.
可靠的方法分三步。第一步，正确求导得到 dy/dx。第二步，把指定的 x 值代入 dy/dx，得到数值梯度 m。第三步，使用直线方程 y - y₁ = m(x - x₁)；其中 y₁ 必须把 x 值代回原方程求得，绝不能用导数求。
Questions often combine two or more tasks, such as finding a stationary point and then the tangent at a different point, or finding the normal at the point where the curve crosses an axis. You should therefore practise linking the techniques in a single flow rather than treating each skill in isolation.
题目经常把两个或多个任务组合起来，例如先求驻点，再求另一点的切线；或者求曲线与坐标轴交点处的法线。因此你应该练习把各个技巧串联成一条完整的解题流程，而不是孤立地练单项技能。
2. Core Differentiation Rules | 核心求导法则
The foundation of the whole exercise is the power rule. If y = xⁿ, then dy/dx = nxⁿ⁻¹. This rule applies term by term when a polynomial is differentiated, and each term keeps its coefficient.
整个练习的基础是幂函数求导法则。若 y = xⁿ，则 dy/dx = nxⁿ⁻¹。对多项式逐项求导时，该法则逐项适用，每项的系数保持不变。
If y = xⁿ then dy/dx = nxⁿ⁻¹
The derivative of a constant is zero, and a constant multiplier stays outside the derivative. For example, if y = 5x³, then dy/dx = 15x²; if y = 7, then dy/dx = 0; if y = 4x, then dy/dx = 4.
常数的导数为零，常数系数在求导时保持不变。例如，若 y = 5x³，则 dy/dx = 15x²；若 y = 7，则 dy/dx = 0；若 y = 4x，则 dy/dx = 4。
- d/dx [xⁿ] = nxⁿ⁻¹ — power rule / 幂函数求导法则
- d/dx [k] = 0 — the derivative of a constant is zero / 常数求导为零
- d/dx [c f(x)] = c f '(x) — constant multiplier is preserved / 常数系数保留
- d/dx [f(x) ± g(x)] = f '(x) ± g'(x) — differentiate term by term / 逐项求导
You should also be able to switch notation without hesitation between dy/dx and f '(x), and between the second derivative d²y/dx² and f ''(x). AQA mark schemes accept either, provided the meaning is clear.
你还应能在 dy/dx 与 f '(x) 之间、以及二阶导数 d²y/dx² 与 f ''(x) 之间毫无迟疑地切换。只要含义清楚，AQA 评分标准接受任意一种写法。
3. Rewriting Roots and Reciprocals as Powers | 把根式与倒数改写成幂
Before differentiating a term such as 4√x or 3/x², it must first be written as a power of x. This single step converts an unfamiliar expression into one the power rule handles directly, and it is one of the most common places where marks are dropped in Exercise 6H.
对 4√x 或 3/x² 这样的项求导之前，必须先把它写成 x 的幂。这一步能把陌生的表达式变成幂函数法则可直接处理的形式，也是练习 6H 中最常见的失分点之一。
- √x = x^½, so d/dx [√x] = ½x^(−½) / 即 √x = x^½，其导数为 ½x^(−½)
- 1/x = x⁻¹, so d/dx [1/x] = −x⁻² / 即 1/x = x⁻¹，其导数为 −x⁻²
- 1/x² = x⁻², so d/dx [1/x²] = −2x⁻³ / 即 1/x² = x⁻²，其导数为 −2x⁻³
- ∛x = x^(1/3), so its derivative is (1/3)x^(−2/3) / 即 ∛x = x^(1/3)，其导数为 (1/3)x^(−2/3)
Worked cases make the pattern clear. For y = 4√x = 4x^½, the derivative is 4 × ½x^(−½) = 2x^(−½) = 2/√x. For y = 3/x² = 3x⁻², the derivative is 3 × (−2)x⁻³ = −6x⁻³ = −6/x³.
具体例子能看清规律。对 y = 4√x = 4x^½，导数为 4 × ½x^(−½) = 2x^(−½) = 2/√x。对 y = 3/x² = 3x⁻²，导数为 3 × (−2)x⁻³ = −6x⁻³ = −6/x³。
A term such as (2x + 1)/x should also be split before differentiating: it becomes 2 + x⁻¹, whose derivative is −x⁻². Dividing each term of the numerator by the denominator is usually safer and faster than attempting the quotient rule, which is not required for Exercise 6H.
像 (2x + 1)/x 这样的项也应在求导前拆分：它化为 2 + x⁻¹，导数为 −x⁻²。把分子各项分别除以分母，通常比使用商法则更安全也更快，而练习 6H 并不要求商法则。
4. Finding the Gradient at a Given Point | 求给定点处的梯度
Once dy/dx is obtained, the gradient of the curve at a particular point is found by substituting the x-coordinate of that point into the derivative. The result is a number, and this number is also the gradient of the tangent at that point.
得到 dy/dx 后，曲线在某一点处的梯度就是把该点的 x 坐标代入导数。结果是一个数值，这个数值同时也是该点处切线的梯度。
For example, if y = x² + 3x, then dy/dx = 2x + 3. At x = 2 the gradient is 2(2) + 3 = 7. For a cubic such as y = x³ − 4x² + 2, dy/dx = 3x² − 8x, and at x = 3 the gradient is 3(9) − 8(3) = 27 − 24 = 3.
例如，若 y = x² + 3x，则 dy/dx = 2x + 3。在 x = 2 处，梯度为 2(2) + 3 = 7。对三次函数 y = x³ − 4x² + 2，dy/dx = 3x² − 8x，在 x = 3 处梯度为 3(9) − 8(3) = 27 − 24 = 3。
7 gives the full coordinates. Confusing the two substitutions is a classic error in this exercise.
点的 y 坐标要单独用曲线原方程求得。例如 y = 2(2)² + 3(2) = 14，或 y = (3)³ − 4(3)² + 2 = 27 − 36 + 2 = −7，从而得到完整坐标。混淆两次代入是本节最典型的错误。
5. Equation of a Tangent | 切线方程
A tangent is a straight line that touches the curve at the given point and has exactly the gradient of the curve there. With gradient m and point (x₁, y₁), its equation is written using the standard straight-line form.
切线是在给定点与曲线相触、且梯度正好等于该点曲线梯度的直线。已知梯度 m 和点 (x₁, y₁)，用标准直线形式即可写出其方程。
y − y₁ = m(x − x₁)
For instance, take y = x² + 3x at the point where x = 2. The y-coordinate is 10, and the gradient is 7, so the tangent is y − 10 = 7(x − 2), which simplifies to y = 7x − 4. Leaving the answer in either form is normally accepted, but AQA examiners prefer a simplified form when the question asks for the equation of the tangent.
例如，对 y = x² + 3x 在 x = 2 处，y 坐标为 10，梯度为 7，所以切线为 y − 10 = 7(x − 2)，化简得 y = 7x − 4。两种形式一般都可接受，但当题目要求"求切线方程"时，AQA 阅卷更偏好化简后的形式。
Always check the final line by substituting the point back in. If x = 2 gives y gent as required. This ten-second check catches most arithmetic slips before they cost marks.
最后一定要把该点代回验证。若 x = 2 时 y = 7(2) − 4 = 10，说明该点确实在切线上。这个十秒钟的检查能在丢分之前抓住大部分算术失误。
6. Equation of a Normal | 法线方程
The normal is the straight line through the point that is perpendicular to the tangent. Because perpendicular lines have gradients whose product is −1, the normal gradient is the negative reciprocal of the tangent gradient.
法线是过该点且与切线垂直的直线。由于两条互相垂直的直线梯度之积为 −1，法线的梯度就是切线梯度的负倒数。
mnormal = −1 / mtangent, and mtangent × mnormal = −1
Continuing the example with tangent gradient 7 at the point (2, 10), the normal gradient is −1/7. The normal therefore satisfies y − 10 = −(1/7)(x − 2), which rearranges to x + 7y = 72 or equivalently y = −x/7 + 72/7.
继续上一个例子：在点 (2, 10) 处切线梯度为 7，法线梯度为 −1/7。因此法线满足 y − 10 = −(1/7)(x − 2)，整理得 x + 7y = 72，或等价地 y = −x/7 + 72/7。
Two special cases deserve attention. If the tangent is horizontal, then m = 0 and the normal is vertical, written x = x₁. If the tangent is vertical, which can occur at a sharp point or a cusp, the normal is horizontal. At A-level these cases are rare but they do appear in stretch questions.
有两种特殊情况需要注意。若切线水平，则 m = 0，法线竖直，写作 x = x₁。若切线竖直（可能出现在尖点或尖角处），法线水平。A-level 中这两种情况较少见，但在拓展题中会出现。
7. Stationary Points: Definition and Conditions | 驻点的定义与条件
A stationary point is a point on the curve where the tangent is horizontal, that is, where the gradient is zero. Since the gradient is given by the derivative, the condition for a stationary point is a single equation to solve.
驻点是曲线上切线水平的点，即梯度为零的点。由于梯度由导数给出，因此驻点的条件就是一个要求解的方程。
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