📚 IB Math AA: Exercise 6E.2 – Tangents, Normals, and Differentiation Techniques | IB数学AA：练习6E.2 – 切线、法线与微分技巧
Exercise 6E.2 in the IB Mathematics: Analysis and Approaches course is a pivotal set of problems designed to deepen your understanding of how differentiation connects algebraic expressions to geometric representations. These exercises typically ask you to find the equations of tangents and normals to curves, often requiring a blend of derivative rules, algebraic manipulation, and coordinate geometry. Mastery of these techniques not only prepares you for exam questions but also builds a strong foundation for further topics such as optimisation and differential equations.
在IB数学：分析与方法课程中，练习6E.2是一组关键的习题，旨在加深你对微分如何将代数表达式与几何表示联系起来的理解。这些习题通常会要求你求出曲线的切线和法线方程，往往需要综合运用导数法则、代数运算和坐标几何知识。熟练掌握这些技巧不仅能帮助你应对考试问题，还能为优化和微分方程等后续主题打下坚实基础。
1. Understanding the Scope of Exercise 6E.2 | 理解练习6E.2的范围
Exercise 6E.2 covers a wide spectrum of differentiation applications. You will work with explicit functions like polynomials, trigonometric functions, exponentials, and logarithms. Some questions also involve composite functions that require the chain rule, while others use implicit differentiation or parametric equations to define curves. The core objective is to determine the gradient at a specific point and then construct the line that just touches the curve (tangent) or the line perpendicular to it (normal).
练习6E.2涵盖了微分应用的广泛范围。你将处理显式函数，如多项式、三角函数、指数函数和对数函数。有些题目还涉及需要使用链式法则的复合函数，而另一些则使用隐函数微分或参数方程来定义曲线。核心目标是求出特定点处的斜率，然后构造恰好接触曲线的直线（切线）或与其垂直的直线（法线）。
2. Essential Differentiation Rules | 基本微分法则
Before tackling tangents and normals, you must be confident with the fundamental differentiation rules. The table below summarises the key rules you will apply in Exercise 6E.2.
| Rule | Form | Derivative Example |
|---|---|---|
| Power Rule | xⁿ | n xⁿ⁻¹ |
| Sum/Difference | f ± g | f' ± g' |
| Product Rule | uv | u'v + uv' |
| Quotient Rule | u/v | (u'v - uv')/v² |
| Chain Rule | f(g(x)) | f'(g(x))·g'(x) |
在着手处理切线与法线之前，你必须熟练掌握基本的微分法则。上表总结了你在练习6E.2中会应用到的关键法则。掌握这些法则后，你将能高效地对多项式、根式、三角函数以及复合函数进行求导，为后续求切线斜率做好准备。
3. Evaluating the Derivative at a Given Point | 计算给定点的导数值
eck that the point actually lies on the curve by plugging the x-value into the original function.
一旦求出一般导数 dy/dx，下一步就是将切点的 x 坐标代入该表达式。这将给出曲线在该点的斜率 m。例如，若 y = x³ - 2x，则 dy/dx = 3x² - 2。在 x = 1 处，斜率为 m = 3(1)² - 2 = 1。请务必通过将 x 值代入原函数来再次确认该点确实位于曲线上。
4. Formulating the Equation of a Tangent | 建立切线方程
ays express the final equation in the requested form, often ax + by + c = 0 or y = mx + c.
Equation of tangent: y - y₁ = m (x - x₁)
已知斜率 m 和点 (x₁, y₁)，就可以使用点斜式构造切线方程。这个公式简单却十分强大。最后一定要将方程写成题目要求的形式，通常是 ax + by + c = 0 或 y = mx + c。
5. Formulating the Equation of a Normal | 建立法线方程
the point of contact. Therefore, its gradient mₙ is the negative reciprocal of the tangent' If the tangent is vertical, the normal is horizontal. The standard formula is shown below.
Equation of normal: y - y₁ = -1/m (x - x₁), m ≠ 0
法线在切点处与切线垂直。因此，如果 m ≠ 0，其斜率 mₙ 是切线斜率 m 的负倒数。若 m = 0，则切线水平，法线竖直（x = x₁）。若切线竖直，则法线水平。标准公式如上所示。
6. Applying the Chain Rule in Tangent Problems | 在切线问题中应用链式法则
0(2x+1)⁴. Once the derivative is found, evaluating it at the required x-coordinate and proceeding with the tangent/normal equation becomes straightforward.
练习6E.2中的许多函数是复合函数，例如 y = (2x + 1)⁵ 或 y = sin³x。对这些函数求导必须使用链式法则：对外层函数求导，然后乘以内层函数的导数。对于 y = (2x+1)⁵，dy/dx = 5(2x+1)⁴·2 = 10(2x+1)⁴。一旦求出导数，在所需 x 坐标处求值并继续求切线/法线方程就变得简单明了。
7. Handling Implicit Differentiation | 处理隐函数微分
n solve for dy/dx algebraically. For the circle x² + y² = 25, implicit differentiation gives 2x + 2y·dy/dx = 0, so dy/dx = -x/y. At the point (3,4), the gradient is -3/4, and the tangent is y - 4 = -3/4 (x - 3).
当曲线由隐式方程定义时，例如 x² + y² = 25 或 x³ + y³ = 6xy，不能简单地在求导前解出 y。应对等式两边关于 x 求导，将 y 视为 x 的函数，并对含有 y 的项使用链式法则（这会生成 dy/dx 因子）。然后通过代数方法解出 dy/dx。对于圆 x² + y² = 25，隐函数微分得 2x + 2y·dy/dx = 0，所以 dy/dx = -x/y。在点 (3,4) 处，斜率为 -3/4，切线为 y - 4 = -3/4 (x - 3)。
8. Tangents and Normals for Parametric Curves | 参数曲线的切线与法线
vided dx/dt ≠ 0. To find the tangent at a specific t-value, first compute the coordinates (x₁, y₁) by plugging t into f and g, then find the gradient via the ratio of derivatives. From there, the tangent and normal equations follow the same point-slope pattern.
dy/dx = (dy/dt) ÷ (dx/dt)
如果曲线由参数方程 x = f(t)、y = g(t) 给出，则斜率 dy/dx 是通过将 y 对 t 的导数除以 x 对 t 的导数得到的。公式为 dy/dx = (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0。要找到特定 t 值处的切线，首先将 t 代入 f 和 g 计算坐标 (x₁, y₁)，然后通过导数之比求出斜率。之后，切线和法线方程遵循相同的点斜式模式。
9. Step-by-Step Worked Example | 分步解题示例
Let' es the product rule, chain rule, and tangent/normal equations. Consider the curve y = x² eˣ where x = 1. We will find the equations of both the tangent and the normal at this point.
我们一起来解一道典型的练习6E.2题目，它综合了乘积法则、链式法则以及切线/法线方程。考虑曲线 y = x² eˣ 在 x = 1 处的点。我们将求出该点处的切线和法线方程。
Step 1: Differentiate using the product rule. Let u = x² and v = eˣ. Then u' = 2x and v' = eˣ. The derivative is dy/dx = u'v + uv' = 2x eˣ + x² eˣ = eˣ (2x + x²).
第一步：使用乘积法则求导。令 u = x²，v = eˣ，则 u' = 2x，v' = eˣ。导数为 dy/dx = u'v + uv' = 2x eˣ + x² eˣ = eˣ (2x + x²)。
Step 2: Find the y-coordinate. When x = 1, y = (1)² e¹ = e. So the point is (1, e).
第二步：求 y 坐标。当 x = 1 时，y = (1)² e¹ = e。因此点为 (1, e)。
Step 3: Calculate the gradient at x = 1. Substitute x = 1 into dy/dx: m = e¹ (2·1 + 1²) = 3e.
第三步
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