Alevel数学 微分法则 链式求导 隐函数求导

Alevel数学 微分法则 链式求导 隐函数求导

Introduction to Differentiation 微分导论

Differentiation is one of the two central pillars of calculus, alongside integration. At its core, differentiation measures the instantaneous rate of change of a function ： how quickly the function’s output changes as its input changes. 微分是微积分的两大核心支柱之一，与积分并列。微分的核心是衡量函数的瞬时变化率：即当输入发生变化时，函数输出变化的速度有多快。

In A-Level Mathematics, differentiation forms the foundation for understanding gradients of curves, rates of change in physical systems, and optimisation problems. The derivative of a function f(x) is denoted as f'(x) or dy/dx when y = f(x), and represents the gradient of the tangent line to the curve at any given point. 在A-Level数学中，微分为理解曲线的斜率、物理系统的变化率以及最优化问题奠定了基础。函数f(x)的导数记作f'(x)或当y = f(x)时的dy/dx，它表示曲线在任意给定点处的切线斜率。

Differentiation from First Principles 从第一性原理求导

The formal definition of the derivative stems from the concept of a limit. For any function f(x), the derivative at a point x is defined as: f'(x) = lim(h = 0) [f(x+h) − f(x)] / h. This expression represents the gradient of a chord connecting two points on the curve, and as h approaches zero, the chord becomes the tangent. 导数的正式定义源于极限的概念。对于任意函数f(x)，在点x处的导数定义为：f'(x) = lim(h = 0) [f(x+h) − f(x)] / h。该表达式表示连接曲线上两点的弦的斜率，当h趋近于零时，弦变为切线。

Let us apply first principles to differentiate f(x) = x². We compute f(x+h) = (x+h)² = x² + 2xh + h², then subtract f(x) to obtain 2xh + h². Dividing by h gives 2x + h, and taking the limit as h = 0 yields 2x. This confirms the well-known result that the derivative of x² is 2x. 让我们用第一性原理对f(x) = x²进行求导。我们计算f(x+h) = (x+h)² = x² + 2xh + h²，减去f(x)得到2xh + h²。除以h得到2x + h，取h = 0的极限得到2x。这验证了众所周知的结果：x²的导数是2x。

Exam questions often ask you to differentiate simple functions like x³ or 1/x from first principles. For x³, expand (x+h)³ = x³ + 3x²h + 3xh² + h³, subtract x³, divide by h, and take the limit to obtain 3x². For f(x) = 1/x, compute f(x+h) − f(x) = 1/(x+h) − 1/x = (x − x − h) / [x(x+h)] = −h / [x(x+h)], divide by h, and let h = 0 to get −1/x². 考试题经常要求你用第一性原理对简单函数如x³或1/x进行求导。对于x³，展开(x+h)³ = x³ + 3x²h + 3xh² + h³，减去x³，除以h，取极限得到3x²。对于f(x) = 1/x，计算f(x+h) − f(x) = 1/(x+h) − 1/x = (x − x − h) / [x(x+h)] = −h / [x(x+h)]，除以h，令h = 0得到−1/x²。

Basic Differentiation Rules 基本求导法则

The Power Rule is the most fundamental differentiation result: if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. This rule applies for any real exponent n, including negative and fractional powers. For example, the derivative of x⁵ is 5x⁴; the derivative of 1/x³ = x⁻³ is −3x⁻⁴; and the derivative of √x = x¹/² is (1/2)x⁻¹/². 幂法则是最基本的微分结果：如果f(x) = xⁿ，则f'(x) = nxⁿ⁻¹。该法则适用于任何实数指数n，包括负指数和分数指数。例如，x⁵的导数是5x⁴；1/x³ = x⁻³的导数是−3x⁻⁴；√x = x¹/²的导数是(1/2)x⁻¹/²。

The Constant Multiple Rule states that the derivative of a constant times a function equals the constant times the derivative of the function: d/dx [c · f(x)] = c · f'(x). For instance, d/dx [7x⁴] = 7 · 4x³ = 28x³. The Sum Rule allows us to differentiate term by term: d/dx [f(x) + g(x)] = f'(x) + g'(x). The Constant Rule states that the derivative of any constant is zero, since a horizontal line has zero gradient everywhere. 常数倍法则指出，常数乘以函数的导数等于常数乘以函数的导数：d/dx [c · f(x)] = c · f'(x)。例如，d/dx [7x⁴] = 7 · 4x³ = 28x³。求和法则允许我们逐项求导：d/dx [f(x) + g(x)] = f'(x) + g'(x)。常数法则指出任何常数的导数为零，因为水平线处处斜率为零。

The Chain Rule 链式法则

The Chain Rule is used to differentiate composite functions ： functions of functions. If y = f(u) and u = g(x), then dy/dx = dy/du · du/dx. In Leibniz notation this is elegantly expressed as the product of two derivatives, where the intermediate variable u effectively cancels out. 链式法则用于对复合函数（函数的函数）进行求导。如果y = f(u)且u = g(x)，则dy/dx = dy/du · du/dx。在莱布尼茨记法中，这被优雅地表示为两个导数的乘积，其中中间变量u被有效地约去。

Consider the function y = (3x² + 5)⁴. Set u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x. Multiplying: dy/dx = 4u³ · 6x = 24x(3x² + 5)³. In practice, experienced mathematicians apply the chain rule mentally: derivative of outer function × derivative of inner function. For y = sin(2x), the outer function is sin(u) with derivative cos(u), and the inner function is 2x with derivative 2, giving dy/dx = 2cos(2x). 考虑函数y = (3x² + 5)⁴。设u = 3x² + 5，则y = u⁴。那么dy/du = 4u³，du/dx = 6x。相乘：dy/dx = 4u³ · 6x = 24x(3x² + 5)³。在实践中，经验丰富的数学家会心算链式法则：外层函数的导数乘以内层函数的导数。对于y = sin(2x)，外层函数是sin(u)，导数为cos(u)，内层函数是2x，导数为2，得到dy/dx = 2cos(2x)。

The Chain Rule extends naturally to deeper nesting. To differentiate y = e^(sin(x²)), work outward: the outermost layer is e^u with derivative e^u; the next layer sin(v) gives cos(v); the innermost x² gives 2x. Thus dy/dx = e^(sin(x²)) · cos(x²) · 2x. A common exam mistake is stopping after one step ： always trace back to the innermost independent variable. 链式法则自然地扩展到更深的嵌套。要对y = e^(sin(x²))求导，从外向内处理：最外层是e^u，导数为e^u；下一层sin(v)给出cos(v)；最内层x²给出2x。因此dy/dx = e^(sin(x²)) · cos(x²) · 2x。一个常见的考试错误是在一步后就停止：一定要追溯到最内层的自变量。

The Product Rule 乘积法则

When a function is expressed as the product of two functions, the Product Rule applies: if y = u(x) · v(x), then dy/dx = u · dv/dx + v · du/dx. In words: the first function times the derivative of the second, plus the second function times the derivative of the first. 当函数表示为两个函数的乘积时，使用乘积法则：如果y = u(x) · v(x)，则dy/dx = u · dv/dx + v · du/dx。简而言之：第一个函数乘以第二个函数的导数，加上第二个函数乘以第一个函数的导数。

Differentiate y = x² · sin(x). Let u = x² with u’ = 2x, and v = sin(x) with v’ = cos(x). Applying the Product Rule: dy/dx = x² · cos(x) + sin(x) · 2x = x[x · cos(x) + 2sin(x)]. A useful verification is to check that the rule is symmetric in u and v, reflecting the commutative property of multiplication. 对y = x² · sin(x)求导。设u = x²，u’ = 2x；v = sin(x)，v’ = cos(x)。应用乘积法则：dy/dx = x² · cos(x) + sin(x) · 2x = x[x · cos(x) + 2sin(x)]。一个有用的验证是检查该法则对u和v是对称的，反映了乘法的交换性质。

For products of three functions, extend the pattern: d/dx [uvw] = u’vw + uv’w + uvw’. For example, differentiate y = x · e^x · ln(x). Each term in the derivative includes the derivative of exactly one factor multiplied by the other two factors unchanged. This pattern generalises to products of n functions. 对于三个函数的乘积，扩展该模式：d/dx [uvw] = u’vw + uv’w + uvw’。例如，求y = x · e^x · ln(x)的导数。导数中的每一项都恰好包含一个因子的导数，乘以其他两个因子不变。这种模式可以推广到n个函数的乘积。

The Quotient Rule 商法则

The Quotient Rule handles functions expressed as a ratio: if y = u(x) / v(x), then dy/dx = (v · du/dx − u · dv/dx) / v². The numerator is “bottom times derivative of top, minus top times derivative of bottom”, and the denominator is the original denominator squared. 商法则处理表示为比值的函数：如果y = u(x) / v(x)，则dy/dx = (v · du/dx − u · dv/dx) / v²。分子是”分母乘以分子的导数，减去分子乘以分母的导数”，分母是原分母的平方。

Differentiate y = (2x + 1) / (x² − 3). Set u = 2x + 1 with u’ = 2, and v = x² − 3 with v’ = 2x. Then: dy/dx = [(x² − 3)(2) − (2x + 1)(2x)] / (x² − 3)² = [2x² − 6 − 4x² − 2x] / (x² − 3)² = [−2x² − 2x − 6] / (x² − 3)². Always simplify the numerator where possible ： examiners often award marks for correct simplification. 对y = (2x + 1) / (x² − 3)求导。设u = 2x + 1，u’ = 2；v = x² − 3，v’ = 2x。则：dy/dx = [(x² − 3)(2) − (2x + 1)(2x)] / (x² − 3)² = [2x² − 6 − 4x² − 2x] / (x² − 3)² = [−2x² − 2x − 6] / (x² − 3)²。尽可能简化分子：考官通常会对正确的简化给予分数。

Many students memorise the Quotient Rule as a separate formula, but it is actually derivable from the Product Rule and Chain Rule. Rewrite u/v as u · v⁻¹, then apply the Product Rule: dy/dx = u’ · v⁻¹ + u · (−1)v⁻² · v’ = (u’v − uv’)/v². Understanding this derivation reduces the memorisation burden and deepens your grasp of how differentiation rules interconnect. 许多学生将商法则记忆为一个独立的公式，但它实际上可以从乘积法则和链式法则推导出来。将u/v重写为u · v⁻¹，然后应用乘积法则：dy/dx = u’ · v⁻¹ + u · (−1)v⁻² · v’ = (u’v − uv’)/v²。理解这个推导可以减少记忆负担，并加深你对求导法则如何相互关联的掌握。

Implicit Differentiation 隐函数求导

Implicit differentiation is required when y cannot easily be expressed as an explicit function of x. For equations like x² + y² = 25 or e^(xy) + sin(y) = x, we differentiate both sides with respect to x, treating y as a function of x and applying the Chain Rule whenever we differentiate a term involving y. 当y不能轻易地表示为x的显函数时，需要使用隐函数求导。对于像x² + y² = 25或e^(xy) + sin(y) = x这样的方程，我们对两边关于x求导，将y视为x的函数，并在对涉及y的项求导时应用链式法则。

Consider the circle equation x² + y² = 25. Differentiating both sides with respect to x: 2x + 2y · dy/dx = 0. The term y² requires the Chain Rule because y is a function of x: d/dx(y²) = 2y · dy/dx. Solving for dy/dx: dy/dx = −x/y. This makes geometric sense ： the gradient of the tangent at any point on a circle equals the negative reciprocal of the radius vector slope. 考虑圆的方程x² + y² = 25。对两边关于x求导：2x + 2y · dy/dx = 0。y²项需要链式法则，因为y是x的函数：d/dx(y²) = 2y · dy/dx。解出dy/dx：dy/dx = −x/y。这在几何上是有意义的：圆上任意点的切线斜率等于半径向量斜率的负倒数。

For more complex implicit functions, the key is systematically applying the Chain Rule to every occurrence of y. To differentiate e^(xy) + sin(y) = x: the term e^(xy) requires the Product Rule within the Chain Rule ： d/dx[e^(xy)] = e^(xy) · (1 · y + x · dy/dx). The term sin(y) gives cos(y) · dy/dx. The right side x gives 1. Collecting all dy/dx terms on one side and solving yields the derivative. 对于更复杂的隐函数，关键是对每一个出现的y系统性地应用链式法则。对e^(xy) + sin(y) = x求导：e^(xy)项需要在链式法则中使用乘积法则：d/dx[e^(xy)] = e^(xy) · (1 · y + x · dy/dx)。sin(y)项给出cos(y) · dy/dx。右边x给出1。将所有dy/dx项集中到一边并求解得到导数。

Implicit differentiation is particularly powerful for finding tangents and normals to curves defined by relations rather than functions. The gradient expression dy/dx often remains in terms of both x and y, so you must substitute the specific point coordinates to find the numeric gradient at that point. 隐函数求导对于求由关系式而非函数定义的曲线的切线和法线特别有用。梯度表达式dy/dx通常仍然以x和y的形式存在，因此你必须代入特定点的坐标来找到该点处的数值梯度。

Parametric Differentiation 参数方程求导

Parametric equations express x and y separately in terms of a third variable, typically t: x = f(t), y = g(t). The derivative dy/dx is found via the chain rule: dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. This elegantly eliminates the parameter t without needing to solve for y in terms of x explicitly. 参数方程将x和y分别表示为第三个变量（通常是t）的函数：x = f(t)，y = g(t)。导数dy/dx通过链式法则求得：dy/dx = (dy/dt) / (dx/dt)，前提是dx/dt ≠ 0。这优雅地消去了参数t，而无需显式地解出y关于x的表达式。

Example: A curve is defined by x = t² − 1 and y = t³ + 3t. Find the equation of the tangent at the point where t = 2. First, compute dx/dt = 2t and dy/dt = 3t² + 3. Then dy/dx = (3t² + 3) / (2t). At t = 2: dy/dx = (12 + 3) / 4 = 15/4. The point coordinates are x = 4 − 1 = 3 and y = 8 + 6 = 14. Using y − y₁ = m(x − x₁): y − 14 = (15/4)(x − 3). 示例：曲线由x = t² − 1和y = t³ + 3t定义。求t = 2处的切线方程。首先，计算dx/dt = 2t，dy/dt = 3t² + 3。然后dy/dx = (3t² + 3) / (2t)。在t = 2处：dy/dx = (12 + 3) / 4 = 15/4。点坐标为x = 4 − 1 = 3，y = 8 + 6 = 14。使用y − y₁ = m(x − x₁)：y − 14 = (15/4)(x − 3)。

For second derivatives in parametric form, use the chain rule again: d²y/dx² = d/dx(dy/dx) = d/dt(dy/dx) / (dx/dt). First find dy/dx as a function of t, then differentiate that expression with respect to t, and finally divide by dx/dt. This is a common A-Level exam question type that tests layered application of the Chain Rule. 对于参数形式的二阶导数，再次使用链式法则：d²y/dx² = d/dx(dy/dx) = d/dt(dy/dx) / (dx/dt)。首先将dy/dx表示为t的函数，然后对该表达式关于t求导，最后除以dx/dt。这是A-Level考试中常见的题型，测试链式法则的分层应用。

Applications of Differentiation 微分的应用

Differentiation enables finding equations of tangents and normals to curves. Given a curve y = f(x) and a point (x₁, y₁) on it, the tangent line has equation y − y₁ = f'(x₁)(x − x₁). The normal line, being perpendicular, has gradient −1/f'(x₁), giving y − y₁ = (−1/f'(x₁))(x − x₁). Tangents and normals appear frequently in coordinate geometry problems. 微分能够求解曲线的切线和法线方程。给定曲线y = f(x)及其上的一点(x₁, y₁)，切线方程为y − y₁ = f'(x₁)(x − x₁)。法线由于垂直，其斜率为−1/f'(x₁)，方程为y − y₁ = (−1/f'(x₁))(x − x₁)。切线和法线在坐标几何问题中频繁出现。

Stationary points occur where dy/dx = 0, indicating locations where the gradient is momentarily zero: maxima, minima, or points of inflection. The second derivative test classifies them: if d²y/dx² > 0 at a stationary point, it is a local minimum; if d²y/dx² < 0, it is a local maximum; if d²y/dx² = 0, further investigation is needed (the point may be a stationary inflection). 驻点出现在dy/dx = 0处，表示梯度瞬时为零的位置：极大值点、极小值点或拐点。二阶导数检验对其进行分类：如果在驻点处d²y/dx² > 0，则为局部极小值；如果d²y/dx² < 0，则为局部极大值；如果d²y/dx² = 0，则需要进一步研究（该点可能是驻点拐点）。

Optimisation problems are among the most practical applications of differentiation. These involve finding maximum or minimum values of quantities such as area, volume, cost, or profit. The method: express the quantity to be optimised as a function of one variable, differentiate, set the derivative to zero, solve for the variable, and verify the nature of the stationary point using the second derivative test. 最优化问题是微分最实际的应用之一。这些问题涉及求解面积、体积、成本或利润等量的最大值或最小值。方法为：将要优化的量表示为单一变量的函数，求导，令导数为零，解出变量，并使用二阶导数检验验证驻点的性质。

Worked Example: A rectangular box with a square base and an open top must hold 32,000 cm³. Find the dimensions that minimise the surface area. Let the base side be x and height be h. Volume: x²h = 32000, so h = 32000/x². Surface area A = x² + 4xh = x² + 4x(32000/x²) = x² + 128000/x. Differentiate: dA/dx = 2x − 128000/x². Set to zero: 2x = 128000/x², so x³ = 64000, giving x = 40. Second derivative: d²A/dx² = 2 + 256000/x³ > 0 at x = 40, confirming a minimum. Height h = 32000/1600 = 20 cm. Optimal dimensions: 40 cm × 40 cm × 20 cm. 例题：一个底面为正方形、顶部敞开的矩形盒子必须容纳32,000 cm³。求使表面积最小化的尺寸。设底面边长为x，高为h。体积：x²h = 32000，故h = 32000/x²。表面积A = x² + 4xh = x² + 4x(32000/x²) = x² + 128000/x。求导：dA/dx = 2x − 128000/x²。令其为零：2x = 128000/x²，故x³ = 64000，得x = 40。二阶导数：d²A/dx² = 2 + 256000/x³，在x = 40处> 0，确认为极小值。高h = 32000/1600 = 20 cm。最优尺寸：40 cm × 40 cm × 20 cm。

Common Mistakes and Exam Tips 常见错误与考试技巧

One of the most frequent errors occurs when students forget to apply the Chain Rule. If you see sin(3x), the derivative is 3cos(3x), not cos(3x). Every time you differentiate a function of a function, multiply by the derivative of the inner expression. 最常见的错误之一是学生忘记应用链式法则。如果你看到sin(3x)，其导数是3cos(3x)，而不是cos(3x)。每次你对复合函数求导时，都要乘以内部表达式的导数。

Another common pitfall is misapplying the Quotient Rule order. Remember: “bottom derivative of top minus top derivative of bottom” ： not the reverse. The minus sign is directional; reversing the terms changes the sign of your answer entirely. A helpful mnemonic is “Low D-High minus High D-Low over Low Low” where Low = denominator and High = numerator. 另一个常见的陷阱是错误地应用商法则的顺序。记住：”分母乘以分子的导数减去分子乘以分母的导数”：而不是反过来。减号是有方向性的；颠倒两项的顺序会完全改变答案的符号。一个有用的记忆口诀是”低D高减高D低除以低低”，其中低=分母，高=分子。

When differentiating implicitly, students often forget to include dy/dx when differentiating terms containing y. Every y-term produces an extra factor of dy/dx through the Chain Rule. After differentiating, always collect all terms involving dy/dx on one side before factorising and solving. 在隐函数求导时，学生经常在对含有y的项求导时忘记包含dy/dx。每个y项通过链式法则都会产生一个额外的dy/dx因子。求导后，在因式分解和求解之前，始终将所有涉及dy/dx的项集中到一边。

For parametric equations, be careful with the second derivative formula. It is NOT simply (d²y/dt²) / (d²x/dt²). The correct formula is d²y/dx² = d/dt(dy/dx) / (dx/dt), which requires you to first find dy/dx as a function of t, differentiate it with respect to t, and then divide by dx/dt. This layered Chain Rule application is a classic A-Level exam discriminator. 对于参数方程，要小心二阶导数公式。它不仅仅是(d²y/dt²) / (d²x/dt²)。正确的公式是d²y/dx² = d/dt(dy/dx) / (dx/dt)，这要求你首先将dy/dx表示为t的函数，对其关于t求导，然后除以dx/dt。这种分层的链式法则应用是典型的A-Level考试区分题。

Finally, always check the domain of your answer. If a problem asks for the maximum area of a shape with a perimeter constraint, your solution variable must be positive and must not exceed the physical bounds implied by the constraint. Mathematical solutions that violate the physical context should be rejected, even if they satisfy dA/dx = 0 algebraically. 最后，始终检查你答案的定义域。如果一道题要求在周长约束下求形状的最大面积，你的解变量必须为正，且不能超过约束所隐含的物理界限。违反物理背景的数学解应该被拒绝，即使它们在代数上满足dA/dx = 0。

Mastering differentiation requires consistent practice across all rule types. Begin with straightforward polynomial differentiation, progress through composite functions with the Chain Rule, then tackle products, quotients, implicit equations, and parametric forms. Each rule builds on the previous, and fluency in one supports fluency in the next. Regular timed practice under exam conditions builds both accuracy and speed. 掌握微分需要在所有法则类型上进行持续练习。从简单的多项式求导开始，逐步过渡到使用链式法则的复合函数，然后处理乘积、商、隐式方程和参数形式。每个法则都建立在前一个之上，对其中一个的熟练程度支持对下一个的熟练程度。在考试条件下定期限时练习可以同时提高准确性和速度。