📚 Exercise 25F.2: Implicit Differentiation Practice | 练习25F.2：隐函数求导专练
you will be able to differentiate relations like x² + y² = 25 or sin(x + y) = xy with confidence and precision.
在IB数学分析与方法课程中，掌握隐函数求导对于求解 y 无法显式表达为 x 的函数的方程至关重要。本文深入拆解练习25F.2，带领你理解核心概念、解析典型例题，并避开常见误区。学完本课，你将能自信准确地对方程 x² + y² = 25 或 sin(x + y) = xy 这类关系式进行求导。
1. What Is Implicit Differentiation? | 什么是隐函数求导？
stead of solving for y first, we differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule whenever we differentiate a term involving y.
隐函数求导是一种用于形如 F(x, y) = 0 的隐式形式（而非显式 y = f(x)）的求导技巧。我们并不先解出 y，而是直接对方程两边关于 x 求导，将 y 视为 x 的函数，并在对含有 y 的项求导时应用链式法则。
2. When to Use Implicit Differentiation | 何时使用隐函数求导
ecisely on such relations, building your algebraic stamina alongside your calculus skills.
当你无法将 y 单独解出，或者解出的表达式过于复杂时，就应该使用隐函数求导。常见情形包括 x³ + y³ = 6xy、y⁵ + 3x²y² = 5x，或者形如 tan(xy) = x + y 的三角关系。练习25F.2正是围绕这类关系式展开，在锻炼微积分技能的同时提升代数运算的耐受力。
3. Core Steps in Exercise 25F.2 | 练习25F.2的核心步骤
Every implicitly defined relation in this exercise can be tackled using a consistent four-step method. First, differentiate both sides of the equation term by term with respect to x. Second, whenever you differentiate a y-term, multiply by dy/dx. Third, collect all terms containing dy/dx on one side of the equation. Fourth, factor out dy/dx and solve for it algebraically. These steps become automatic with practice.
练习中每个隐式定义的关系都可以用一套四步法解决。第一步，对方程两边逐项关于 x 求导。第二步，每当你对含 y 的项求导时，要乘上 dy/dx。第三步，把所有含 dy/dx 的项集中到方程的一侧。第四步，提取 dy/dx 并通过代数运算解出它。通过练习，这些步骤会变得十分自然。
4. Worked Example 1: Circle Equation | 例题1：圆的方程
= 25. Differentiating both sides gives 2x + 2y·(dy/dx) = 0. Isolating dy/dx yields dy/dx = – he tangent line at any point (x, y) on the circle is the negative reciprocal of the radius's slope, a property unique to circles.
考虑方程 x² + y² = 25。两边求导得到 2x + 2y·(dy/dx) = 0。分离 dy/dx 后得出 dy/dx = –x/y。这一结果说明，圆上任意点 (x, y) 处切线的斜率等于该点半径斜率的负倒数，这是圆特有的几何性质。
d/dx (x² + y²) = d/dx (25) → 2x + 2y·dy/dx = 0 → dy/dx = –x/y
5. Worked Example 2: Product of x and y | 例题2：x与y的乘积
wise: (1·y³ + x·3y²·dy/dx) + (2x·y + x²·dy/dx) = 0. Group dy/dx terms: (3xy² + x²)·dy/dx = –y³ – 2xy. Hence, dy/dx = –(y³ + 2xy)/(3xy² + x²).
当关系式含有 x 与 y 的乘积时，必须细致运用乘法法则。对于 xy³ + x²y = 5，逐项求导得：(1·y³ + x·3y²·dy/dx) + (2x·y + x²·dy/dx) = 0。合并含 dy/dx 的项：(3xy² + x²)·dy/dx = –y³ – 2xy。因此 dy/dx = –(y³ + 2xy)/(3xy² + x²)。
6. Worked Example 3: Trigonometric Relation | 例题3：三角关系式
cos(x + y) + cos(x + y)·dy/dx = y + x·dy/dx. Bring dy/dx terms together: cos(x + y)·dy/dx – x·dy/dx = y – cos(x + y). Factor and solve: dy/dx = [y – cos(x + y)] / [cos(x + y) – x].
对于 sin(x + y) = xy，左边用链式法则：cos(x + y)·(1 + dy/dx) = y + x·dy/dx。展开左边并集中 dy/dx 项：cos(x + y) + cos(x + y)·dy/dx = y + x·dy/dx。将含 dy/dx 的项移至同一侧：cos(x + y)·dy/dx – x·dy/dx = y – cos(x + y)。提取公因式并求解：dy/dx = [y – cos(x + y)] / [cos(x + y) – x]。
7. Finding the Second Derivative Implicitly | 隐函数二阶导数的求法
to x, remembering to apply the chain rule to any y that appears. For x² + y² = 1, dy/dx = –x/y. To find d²y/dx², use the quotient rule: d/dx (–x/y) = –(1·y – x·dy/dx)/y². Substitute dy/dx = –x/y and simplify to obtain d²y/dx² = –(y² + x²)/y³ = –1/y³. Such results test your algebraic manipulation under pressure.
练习25F.2偶尔会要求计算 d²y/dx²。在求出 dy/dx 后，需对其再次关于 x 求导，并牢记对出现的任何 y 使用链式法则。对 x² + y² = 1，已得 dy/dx = –x/y。为求 d²y/dx²，使用商法则：d/dx (–x/y) = –(1·y – x·dy/dx)/y²。代入 dy/dx = –x/y 并化简，得到 d²y/dx² = –(y² + x²)/y³ = –1/y³。这类结果考验的是你在压力下的代数运算能力。
8. Common Mistakes to Avoid | 常见错误及避免方法
One frequent error is forgetting to multiply by dy/dx when differentiating a function of y. Another is misapplying the product rule when both x and y appear multiplicatively. Students also often lose a negative sign when moving terms. Lastly, confusing dy/dx with 1/(dx/dy) can cause mistakes in related rates problems. Slow down, write every step, and check each term before collecting like terms.
一个常见错误是，在对 y 的函数求导时忘记乘 dy/dx。其次是在 x 与 y 相乘出现时误用乘法法则。学生也经常在移项时遗漏负号。最后，混淆 dy/dx 与 1/(dx/dy) 会在相关变化率问题中导致错误。放慢速度，写下每一步，并在合并同类项前逐项检查。
9. Practice Problems from Exercise 25F.2 | 练习25F.2的配套练习
pt the following representative problems. (a) Find dy/dx if x⁴ + 2x²y + y³ = 10. (b) For eʸ = 3x, determine the slope of the tangent line at (0, 0). (c) If y is defined implicitly by √x + y + √xy 4, evaluate dy/dx at (2, 2). (d) Show that the curve x²/³ + y²/³ = a²/³ satisfies dy/dx = –¹/³. Each problem reinforces a different aspect of the implicit differentiation technique.
为打磨技能，请尝试以下代表性题目。(a) 若 x⁴ + 2x²y + y³ = 10，求 dy/dx。(b) 对于 eʸ + cos(xy) = 3x，确定点 (0, 0) 处切线的斜率。(c) 若 y 由 √x + y + √xy = 4 隐式定义，求在 (2, 2) 处的 dy/dx。(d) 证明曲线 x²/³ + y²/³ = a²/³ 满足 dy/dx = –(y/x)¹/³。每道题都强化了隐函数求导技巧的不同侧面。
10. Applications: Related Rates | 应用：相关变化率
nd dx/dt = 2 units/s when x = 3 and y = 4, we find 2x·dx/dt + 2y·dy/dt = 0, giving dy/dt = – ition from implicit differentiation to dynamic scenarios makes the exercise indispensable.
练习25F.2打磨的技能自然延伸至相关变化率问题。当两个变量隐式关联且两者都随时间 t 变化时，对关系式关于 t 求导即会产生 dx/dt 和 dy/dt 项。例如，若 x² + y² = 25，且在 x = 3, y = 4 时 dx/dt = 2 单位/秒，则由 2x·dx/dt + 2y·dy/dt = 0 可求出 dy/dt = –1.5 单位/秒。这种从隐函数求导向动态场景的无缝过渡，使得本练习不可或缺。
11. Checking Your Answers with Technology | 用技术工具验证答案
ts provides a visual check. For example, after finding dy/dx for x² + y² = 25 at (3, 4) as –3/4, plot the circle and the line y – 4 = –3/4 (x – tly one point. Such checks reduce algebraic errors and reinforce your geometric intuition.
通过手算完成练习25F.2固然能培养基本流畅度，但使用图形计算器或GeoGebra等软件绘制隐式曲线及其切线，可提供直观验证。例如，对 x² + y² = 25 在点 (3, 4) 求出 dy/dx = –3/4 后，绘制圆以及直线 y – 4 = –3/4 (x – 3)，确认该直线恰好在一点接触曲线。这种验证方法能减少代数错误，并强化几何直觉。
12. Linking to the IB Syllabus and Exam Tips | 关联IB考纲与应试技巧
In the IB Analysis and Approaches exam, implicit differentiation typically appears in Paper 1 and Paper 2, often within a multi-part question that tests differential equations, tangents, or normal lines. When tackling such items, always begin by stating that you are differentiating with respect to x, write d/dx before each term, and show the dy/dx factor explicitly. Where possible, simplify the final expression by substituting the original relation to achieve a neater form, as graders reward elegant simplification.
在IB分析与方法考试中，隐函数求导通常出现在试卷一和试卷二，往往交织在考查微分方程、切线或法线的多问答题中。解答这类题目时，务必先声明你是在关于 x 求导，在每一项前写上 d/dx，并明确写出 dy/dx 因子。若可能，可通过回代原关系式化简最终表达式，使其更简洁，因为阅卷人会奖励优雅的化简。
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