Leibniz Formula and Its Applications | 莱布尼茨公式及其应用

📚 Leibniz Formula and Its Applications | 莱布尼茨公式及其应用

The Leibniz formula generalizes the product rule to higher-order derivatives. It is a cornerstone of calculus, essential for solving problems involving repeated differentiation of products of functions.

莱布尼茨公式将乘积法则推广到高阶导数，是微积分中的重要工具，广泛应用于处理两个函数乘积的高阶求导问题。

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1. Statement of the Formula | 公式表述

For two functions u(x) and v(x) that are differentiable n times, the n-th derivative of their product is given by the Leibniz formula:

对于两个具有 n 阶导数的函数 u(x) 与 v(x)，它们乘积的 n 阶导数由莱布尼茨公式给出：

(u·v)⁽ⁿ⁾ = Σₖ₌₀ⁿ C(n, k) u^(k) v^(n−k)

Here C(n, k) is the binomial coefficient, and u⁽⁰⁾ = u, v⁽⁰⁾ = v.

其中 C(n, k) 为二项式系数，并规定 u⁽⁰⁾ = u，v⁽⁰⁾ = v。

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2. Connection with the Binomial Theorem | 与二项式定理的联系

The Leibniz formula has exactly the same structure as the binomial theorem. Each term combines a binomial coefficient and derivatives of the two factors whose orders sum to n.

莱布尼茨公式的结构与二项式定理完全相似：每一项都包含一个二项式系数，以及两个因子的导数，其阶数之和恰为 n。

This analogy helps memorize the formula. If we think of the exponent in (a + b)ⁿ as the order of differentiation, the pattern is identical.

这种类比有助于记忆。如果将二项式定理中的指数视为求导阶数，两者的展开模式完全一致。

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3. Proof by Mathematical Induction | 用数学归纳法证明

We prove the formula by induction. For n = 1, the statement reduces to the ordinary product rule: (uv)’ = u’v + uv’.

我们用数学归纳法证明。当 n = 1 时，结论就是普通的乘积法则：(uv)’ = u’v + uv’。

Assume the formula holds for n. Differentiate both sides with respect to x, then combine terms using Pascal’s identity C(n, k−1) + C(n, k) = C(n+1, k). This yields the formula for n + 1.

假设公式对 n 成立，对等式两边求导，并利用帕斯卡恒等式 C(n, k−1) + C(n, k) = C(n+1, k) 合并同类项，即可得到对 n + 1 的公式。

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4. Finding High-Order Derivatives at a Point | 求某点处的高阶导数

One powerful application is computing the value of a high-order derivative at a specific point without finding the entire derivative function. For example, let y = x² e^x, find y⁽⁵⁰⁾(0).

莱布尼茨公式的一个重要应用是求函数在特定点处的高阶导数值，而无需写出完整的导函数。例如，设 y = x² e^x，求 y⁽⁵⁰⁾(0)。

Take u = x² and v = e^x. Since u^(k) = 0 for k ≥ 3, only three terms remain. The result is C(50,2)·2!·e⁰ = 2450.

取 u = x²，v = e^x。由于当 k ≥ 3 时 u^(k) = 0，展开式中仅剩三项，计算结果为 C(50,2)·2!·e⁰ = 2450。

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5. Application to Polynomial Times Trigonometric Functions | 多项式乘三角函数的应用

When differentiating expressions like x³ sin x repeatedly, the Leibniz formula neatly separates the polynomial part, whose derivatives eventually vanish.

对于 x³ sin x 这类多项式与三角函数乘积的高阶导数，莱布尼茨公式可以将多项式部分单独处理，而多项式的导数最终会变为零，从而简化计算。

For instance, if y = x³ sin x, then y⁽¹⁰⁾ = C(10,0)x³ sin⁽¹⁰⁾x + C(10,1)(x³)’ sin⁽⁹⁾x + … + C(10,3)(x³)”’ sin⁽⁷⁾x. All later terms vanish because (x³)⁽⁴⁾ = 0.

例如，若 y = x³ sin x，则 y⁽¹⁰⁾ = C(10,0)x³ sin⁽¹⁰⁾x + C(10,1)(x³)’ sin⁽⁹⁾x + … + C(10,3)(x³)”’ sin⁽⁷⁾x。由于 (x³)⁽⁴⁾ = 0，后面的项均为零。

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6. Using the Formula for Inverse Tangent | 对反正切函数的高阶导数

Leibniz formula can help derive recurrence relations. For f(x) = arctan x, note that f'(x) = 1/(1 + x²), so (1 + x²)f'(x) = 1.

莱布尼茨公式可用于推导递推关系。设 f(x) = arctan x，因为 f'(x) = 1/(1 + x²)，所以 (1 + x²)f'(x) = 1。

Differentiating both sides n times using Leibniz gives (1 + x²)f⁽ⁿ⁺¹⁾(x) + 2n x f⁽ⁿ⁾(x) + n(n−1) f⁽ⁿ⁻¹⁾(x) = 0, a recurrence for the derivatives at x = 0.

对等式两边用莱布尼茨公式求 n 阶导数，得到 (1 + x²)f⁽ⁿ⁺¹⁾(x) + 2n x f⁽ⁿ⁾(x) + n(n−1) f⁽ⁿ⁻¹⁾(x) = 0，这就是 f⁽ⁿ⁾(0) 的递推关系。

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7. Repeated Product of Three Functions | 三个函数乘积的情形

The Leibniz formula can be extended to the product of several functions. For three functions u, v, w, the n-th derivative expands as a multinomial sum:

莱布尼茨公式可以推广到多个函数乘积的情形。对于三个函数 u、v、w，其 n 阶导数按多重求和展开：

(uvw)⁽ⁿ⁾ = Σ_{i+j+k=n} [n!/(i! j! k!)] u⁽ⁱ⁾ v^(j) w^(k)

This generalized form is useful in physics and engineering when dealing with products of several time-dependent quantities.

这种推广形式在物理和工程中处理多个随时间变化量乘积时非常有用。

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8. Solving Initial Value Problems via Taylor Series | 用泰勒级数求解初值问题

Leibniz formula is often used to generate Taylor coefficients of a solution to a differential equation with given initial conditions.

莱布尼茨公式常被用来生成满足初始条件的微分方程解的泰勒系数。

For example, for y” + x y = 0 with y(0) = 1, y'(0) = 0, differentiate the equation repeatedly. At each step, Leibniz formula handles the term x·y, giving y⁽ⁿ⁺²⁾(0) = −n y⁽ⁿ⁻¹⁾(0). This yields a recurrence for the series coefficients.

例如，对于 y” + x y = 0 且 y(0) = 1，y'(0) = 0，反复对原方程求导。每一步用莱布尼茨公式处理 x·y 项，得到 y⁽ⁿ⁺²⁾(0) = −n y⁽ⁿ⁻¹⁾(0)，从而获得级数系数的递推公式。

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9. Example: y = x ln(1 + x) | 例题：y = x ln(1 + x)

Let y = x ln(1 + x). Find y⁽¹⁰⁰⁾(x). Use Leibniz with u = x, v = ln(1+x). Since u’ = 1 and u” = 0, only two terms remain:

设 y = x ln(1 + x)，求 y⁽¹⁰⁰⁾(x)。取 u = x，v = ln(1+x)，因为 u’ = 1，u” = 0，所以仅剩两项：

y⁽¹⁰⁰⁾ = x·[ln(1+x)]⁽¹⁰⁰⁾ + 100·[ln(1+x)]⁽⁹⁹⁾

Using the known derivative of ln(1+x), namely [ln(1+x)]^(m) = (−1)^m−1(m−1)!/(1+x)^m, we obtain the final result.

利用 ln(1+x) 的已知 m 阶导数 [ln(1+x)]^(m) = (−1)^m−1(m−1)!/(1+x)^m，即可得到最终结果。

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10. Common Pitfalls and Exam Tips | 常见错误与考试要点

A common mistake is to forget the binomial coefficients or to misplace the order of derivatives. Always write out the first few terms explicitly before simplifying.

常见错误包括遗漏二项式系数或颠倒导数的阶数。考试中应先写出展开的前几项，再化简，避免出错。

When one factor is a polynomial of degree m, only terms up to u^(m) survive. This drastically reduces the number of terms and is often the key to solving exam problems.

若其中一个因子是次数为 m 的多项式，则只有直到 u^(m) 的项保留，其余均为零。这能大幅减少项数，往往是解题的关键。

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