📚 Laplace Transforms of Derivatives of Functions | 函数导数的拉普拉斯变换
The Laplace transform is a powerful integral transform used extensively in engineering and physics to convert differential equations into algebraic equations. Among its most important properties is the transform of derivatives, which directly links the original function’s behavior to its rate of change.
拉普拉斯变换是一种强大的积分变换，广泛应用于工程和物理学中，将微分方程转化为代数方程。其中最重要的性质之一就是导数的变换，它直接将原函数的行为与其变化率联系起来。
For a function f(t) defined for t ≥ 0, the Laplace transform is defined as F(s) = L{f(t)} = ∫₀^∞ e^(-st) f(t) dt, provided the integral converges. The transform of derivatives allows us to incorporate initial conditions naturally into the algebraic framework.
对于定义在 t ≥ 0 上的函数 f(t)，拉普拉斯变换定义为 F(s) = L{f(t)} = ∫₀^∞ e^(-st) f(t) dt，前提是积分收敛。导数的变换使我们能够将初始条件自然地纳入代数框架中。
1. Definition and Notation | 定义与记号
Before exploring derivatives, we must fix our notation. Let f(t) be a piecewise continuous function of exponential order, meaning there exist constants M > 0 and a such that |f(t)| ≤ M e^(at) for sufficiently large t. Its Laplace transform is denoted by L{f(t)} = F(s).
在探讨导数之前，我们必须固定记号。设 f(t) 是一个分段连续的指数阶函数，即存在常数 M > 0 和 a，使得对于足够大的 t 有 |f(t)| ≤ M e^(at)。其拉普拉斯变换记为 L{f(t)} = F(s)。
We denote the first derivative as f'(t) = df/dt, the second derivative as f”(t) = d²f/dt², and in general the n-th derivative as f⁽ⁿ⁾(t) = dⁿf/dtⁿ. The initial values f(0), f'(0), f”(0), etc., are evaluated at t = 0⁺ (from the right).
我们记一阶导数为 f'(t) = df/dt，二阶导数为 f”(t) = d²f/dt²，一般地，n 阶导数为 f⁽ⁿ⁾(t) = dⁿf/dtⁿ。初始值 f(0)、f'(0)、f”(0) 等均在 t = 0⁺（从右侧）处取值。
For the Laplace transform to be applicable to derivatives, we require that f and its derivatives satisfy the same convergence conditions. This is usually guaranteed in standard IB problems involving exponential, polynomial, sinusoidal, and combination functions.
为使拉普拉斯变换适用于导数，我们要求 f 及其导数满足相同的收敛条件。在 IB 标准问题中，涉及指数函数、多项式、正弦函数及其组合时，这一条件通常得到满足。
2. Laplace Transform of the First Derivative | 一阶导数的拉普拉斯变换
The fundamental formula for the first derivative is:
一阶导数的基本公式为：
L{f'(t)} = s F(s) – f(0)
This result states that differentiating in the time domain corresponds to multiplying by s in the Laplace domain, together with a correction term containing the initial value f(0). The correction term is essential; without it, the formula would be incomplete.
该结果表明，在时域中求导对应于在拉普拉斯域中乘以 s，同时加上包含初始值 f(0) 的修正项。修正项至关重要；缺少它，公式将不完整。
Let us verify with a simple example. Take f(t) = e^(3t). Then f'(t) = 3e^(3t), and L{f(t)} = 1/(s – 3). Applying the formula: L{f'(t)} = s · (1/(s – 3)) – 1 = (s – (s – 3))/(s – 3) = 3/(s – 3). This matches L{3e^(3t)} directly, confirming the formula.
让我们用一个简单例子验证。取 f(t) = e^(3t)。则 f'(t) = 3e^(3t)，且 L{f(t)} = 1/(s – 3)。代入公式：L{f'(t)} = s · (1/(s – 3)) – 1 = (s – (s – 3))/(s – 3) = 3/(s – 3)。这与直接计算 L{3e^(3t)} 一致，验证了公式。
Another illustrative case is f(t) = sin(2t). Then L{f(t)} = 2/(s² + 4), f(0) = 0, and f'(t) = 2cos(2t). By the formula, L{f'(t)} = s · (2/(s² + 4)) – 0 = 2s/(s² + 4). Indeed, L{2cos(2t)} = 2s/(s² + 4), which is correct.
另一个说明性例子是 f(t) = sin(2t)。则 L{f(t)} = 2/(s² + 4)，f(0) = 0，且 f'(t) = 2cos(2t)。根据公式，L{f'(t)} = s · (2/(s² + 4)) – 0 = 2s/(s² + 4)。确实，L{2cos(2t)} = 2s/(s² + 4)，正确无误。
3. Laplace Transform of the Second Derivative | 二阶导数的拉普拉斯变换
Applying the first-derivative formula twice yields the transform of the second derivative:
对一阶导数公式使用两次，得到二阶导数的变换：
L{f”(t)} = s² F(s) – s f(0) – f'(0)
The pattern is clear: each derivative introduces an additional power of s multiplying F(s), and the initial conditions appear in descending powers of s. Note that the last term is f'(0), not multiplied by s.
规律很明显：每求一次导，F(s) 前就多乘一个 s 的幂，而初始条件以 s 的降幂形式出现。注意最后一项是 f'(0)，不乘以 s。
Let us test with f(t) = t². We know f(0) = 0, f'(0) = 0, f”(t) = 2, and L{t²} = 2/s³. Then the formula gives L{f”(t)} = s² · (2/s³) – s · 0 – 0 = 2/s. Since L{2} = 2/s, the result is correct.
让我们以 f(t) = t² 检验。已知 f(0) = 0，f'(0) = 0，f”(t) = 2，且 L{t²} = 2/s³。则公式给出 L{f”(t)} = s² · (2/s³) – s · 0 – 0 = 2/s。由于 L{2} = 2/s，结果正确。
A more involved example: take f(t) = cos(ωt). Then f(0) = 1, f'(0) = 0, f”(t) = -ω² cos(ωt), and L{f(t)} = s/(s² + ω²). Applying the formula: L{f”(t)} = s² · (s/(s² + ω²)) – s · 1 – 0 = s³/(s² + ω²) – s = (s³ – s(s² + ω²))/(s² + ω²) = -sω²/(s² + ω²). This equals -ω² · (s/(s² + ω²)) = L{-ω² cos(ωt)}, as expected.
一个更复杂的例子：取 f(t) = cos(ωt)。则 f(0) = 1，f'(0) = 0，f”(t) = -ω² cos(ωt)，且 L{f(t)} = s/(s² + ω²)。代入公式：L{f”(t)} = s² · (s/(s² + ω²)) – s · 1 – 0 = s³/(s² + ω²) – s = (s³ – s(s² + ω²))/(s² + ω²) = -sω²/(s² + ω²)。这等于 -ω² · (s/(s² + ω²)) = L{-ω² cos(ωt)}，与预期一致。
4. Higher-Order Derivatives | 高阶导数
For the n-th derivative, the general formula is:
对于 n 阶导数，一般公式为：
L{f⁽ⁿ⁾(t)} = sⁿ F(s) – sⁿ⁻¹ f(0) – sⁿ⁻² f'(0) – … – s f⁽ⁿ⁻²⁾(0) – f⁽ⁿ⁻¹⁾(0)
In compact summation notation: L{f⁽ⁿ⁾} = sⁿ F(s) – Σ_{k=0}^{n-1} sⁿ⁻¹⁻ᵏ f⁽ᵏ⁾(0). Each initial condition of order k is multiplied by s raised to the power (n – 1 – k).
用紧凑的求和记号表示：L{f⁽ⁿ⁾} = sⁿ F(s) – Σ_{k=0}^{n-1} sⁿ⁻¹⁻ᵏ f⁽ᵏ⁾(0)。每个 k 阶初始条件乘以 s 的 (n – 1 – k) 次幂。
For the third derivative, for example:
以三阶导数为例：
L{f”'(t)} = s³ F(s) – s² f(0) – s f'(0) – f”(0)
Notice that the coefficients of the initial conditions are exactly the binomial-like sequence of powers s², s, 1. This pattern continues naturally for all orders, making it easy to memorize: subtract the initial values in order, starting with the highest power of s.
注意初始条件的系数恰好是 s²、s、1 这样二项式般的降幂序列。这一模式自然延续到所有阶数，方便记忆：从 s 的最高次幂开始，依次减去各项初始值。
For example, if f(t) = t³, then f(0) = 0, f'(0) = 0, f”(0) = 0, f”'(t) = 6, and L{t³} = 6/s⁴. The formula gives L{f”'(t)} = s³ · (6/s⁴) – 0 – 0 – 0 = 6/s, which equals L{6} as required.
例如，若 f(t) = t³，则 f(0) = 0，f'(0) = 0，f”(0) = 0，f”'(t) = 6，且 L{t³} = 6/s⁴。公式给出 L{f”'(t)} = s³ · (6/s⁴) – 0 – 0 – 0 = 6/s，等于 L{6}，符合要求。
5. Derivation of the First-Derivative Formula | 一阶导数公式的推导
To understand deeply, we derive the first-derivative formula using integration by parts. Start with the definition: L{f'(t)} = ∫₀^∞ e^(-st) f'(t) dt.
为了深入理解，我们用分部积分法推导一阶导数公式。从定义出发：L{f'(t)} = ∫₀^∞ e^(-st) f'(t) dt。
Let u = e^(-st) and dv = f'(t) dt. Then du = -s e^(-st) dt and v = f(t). Integration by parts yields:
令 u = e^(-st)，dv = f'(t) dt。则 du = -s e^(-st) dt，v = f(t)。分部积分得到：
[e^(-st) f(t)]₀^∞ + s ∫₀^∞ e^(-st) f(t) dt
Assuming f(t) is of exponential order, the boundary term at t → ∞ vanishes because e^(-st) decays faster than f(t) grows for sufficiently large s. At t = 0, the term is -f(0). Therefore the result is s F(s) – f(0).
假设 f(t) 是指数阶函数，当 t → ∞ 时边界项消失，因为对于足够大的 s，e^(-st) 的衰减比 f(t) 的增长更快。在 t = 0 处，该项为 -f(0)。因此结果为 s F(s) – f(0)。
This derivation reveals why the initial condition appears: it comes directly from evaluating the boundary term at the lower limit t = 0. The same logic extends to higher derivatives by repeated application.
这一推导揭示了初始条件为何出现：它直接来自在下限 t = 0 处对边界项的求值。同样的逻辑通过反复应用可推广到高阶导数。
6. Applications to Differential Equations | 在微分方程中的应用
The most significant application of derivative transforms is solving initial value problems. Consider a simple first-order linear ODE: y'(t) + 2y(t) = e^(t), with y(0) = 1.
导数变换最重要的应用是求解初值问题。考虑一个简单的一阶线性常微分方程：y'(t) + 2y(t) = e^(t)，且 y(0) = 1。
Take the Laplace transform of both sides. Using L{y'(t)} = sY(s) – y(0), we obtain: sY(s) – 1 + 2Y(s) = 1/(s – 1). Collecting terms: (s + 2)Y(s) = 1/(s – 1) + 1 = (1 + s – 1)/(s – 1) = s/(s – 1). Thus Y(s) = s/[(s – 1)(s + 2)].
对两边取拉普拉斯变换。利用 L{y'(t)} = sY(s) – y(0)，得到：sY(s) – 1 + 2Y(s) = 1/(s – 1)。整理各项：(s + 2)Y(s) = 1/(s – 1) + 1 = (1 + s – 1)/(s – 1) = s/(s – 1)。因此 Y(s) = s/[(s – 1)(s + 2)]。
Performing partial fraction decomposition: s/[(s – 1)(s + 2)] = A/(s – 1) + B/(s + 2). Solving A + B = 1 and 2A – B = 0 gives A = 1/3, B = 2/3. Thus y(t) = L⁻¹{1/[3(s – 1)]} + L⁻¹{2/[3(s + 2)]} = (1/3)e^(t) + (2/3)e^(-2t).
进行部分分式分解：s/[(s – 1)(s + 2)] = A/(s – 1) + B/(s + 2)。求解 A + B = 1 和 2A – B = 0 得 A = 1/3，B = 2/3。因此 y(t) = L⁻¹{1/[3(s – 1)]} + L⁻¹{2/[3(s + 2)]} = (1/3)e^(t) + (2/3)e^(-2t)。
For a second-order example, take y” + 4y = sin(2t), y(0) = 0, y'(0) = 0. Taking transforms: s²Y – s·0 – 0 + 4Y = 2/(s² + 4). Hence Y(s) = 2/[(s² + 4)²]. Using the known inverse L⁻¹{2a³/[(s² + a²)²]} = sin(at) – at cos(at) with a = 2, we get y(t) = (1/8)sin(2t) – (1/4)t cos(2t).
再举一个二阶例子：y” + 4y = sin(2t)，y(0) = 0，y'(0) = 0。取变换：s²Y – s·0 – 0 + 4Y = 2/(s² + 4)。因此 Y(s) = 2/[(s² + 4)²]。利用已知逆变换 L⁻¹{2a³/[(s² + a²)²]} = sin(at) – at cos(at)，其中 a = 2，得到 y(t) = (1/8)sin(2t) – (1/4)t cos(2t)。
7. Physical Interpretation | 物理意义
In mechanical systems, displacement x(t), velocity x'(t), and acceleration x”(t) are naturally linked. The Laplace transform converts the differential equation of motion m x” + c x’ + k x = f(t) into an algebraic equation: (m s² + c s + k) X(s) = F(s) + m s x₀ + m x’₀ + c x₀.
在机械系统中，位移 x(t)、速度 x'(t) 和加速度 x”(t) 自然相连。拉普拉斯变换将运动微分方程 m x” + c x’ + k x = f(t) 转化为代数方程：(m s² + c s + k) X(s) = F(s) + m s x₀ + m x’₀ + c x₀。其中 x₀ = x(0)，x’₀ = x'(0)。
The term m s x₀ represents the effect of the initial displacement on the system’s Laplace-domain response, while m x’₀ represents the effect of the initial velocity. This separation makes it easy to analyze the contribution of each initial condition.
项 m s x₀ 表示初始位移对系统拉普拉斯域响应的影响，而 m x’₀ 表示初始速度的影响。这种分离便于分析每个初始条件的贡献。
In electrical circuits, the derivative formula mirrors the voltage-current relations of inductors and capacitors. For an inductor with L di/dt = v(t), the transform gives L(s I(s) – i(0)) = V(s), encapsulating both the dynamic behavior and the initial current.
在电路中，导数公式对应于电感和电容的电压-电流关系。对于电感，L di/dt = v(t)，变换后得到 L(s I(s) – i(0)) = V(s)，既体现了动态行为，也包含了初始电流。
8. Common Mistakes and How to Avoid Them | 常见错误与规避方法
One frequent error is forgetting the initial condition terms. Students often write L{f’} = sF(s) without subtracting f(0). This is incorrect and will lead to wrong solutions. Always explicitly write down f(0) before manipulating the equation.
一个常见错误是忘记初始条件项。学生常将 L{f’} = sF(s) 写错，没有减去 f(0)。这是不正确的，会导致错误的解。在操作方程之前，务必明确写出 f(0)。
Another mistake is confusing the signs of the initial condition terms for higher derivatives. For f”, the formula is s²F(s) – s f(0) – f'(0); note that both initial terms are negative. For f”’, all three initial terms are negative: -s²f(0) – s f'(0) – f”(0).
另一个错误是混淆高阶导数中初始条件项的符号。对于 f”，公式为 s²F(s) – s f(0) – f'(0)；注意两个初始项均为负号。对于 f”’，三个初始项均为负号：-s²f(0) – s f'(0) – f”(0)。
Students may also misuse the transform when f or its derivatives are discontinuous at t = 0. The formulas assume right-hand limits at t = 0⁺. If the problem specifies conditions at t = 0⁻, adjustments may be required, though this is rare in IB.
当 f 或其导数在 t = 0 处不连续时，学生可能会误用变换。这些公式假设在 t = 0⁺ 处取右极限。如果问题指定 t = 0⁻ 处的条件，则可能需要进行调整，尽管这在 IB 中很少见。
Finally, be careful when applying the derivative property to inverse transforms. If L⁻¹{F(s)} = f(t), then L⁻¹{s F(s)} = f'(t) + f(0)δ(t) only in a distributional sense; in the context of this course, it is safer to use the forward transform formula directly.
最后，将导数性质应用于逆变换时要小心。如果 L⁻¹{F(s)} = f(t)，则仅当在广义函数意义下 L⁻¹{s F(s)} = f'(t) + f(0)δ(t)；在本课程范围内，直接使用正变换公式更为安全。
9. Table of Key Transform Pairs | 关键变换对一览表
The following table summarizes the most common functions and their Laplace transforms, which are often used together with derivative transforms in problem solving.
下表总结了最常见的函数及其拉普拉斯变换，这些在解题时经常与导数变换一起使用。
| f(t) | F(s) = L{f(t)} |
| 1 | 1/s, s > 0 |
| tⁿ (n = 1, 2, 3, …) | n!/sⁿ⁺¹, s > 0 |
| e^(at) | 1/(s – a), s > a |
| sin(ωt) | ω/(s² + ω²), s > 0 |
| cos(ωt) | s/(s² + ω²), s > 0 |
| t e^(at) | 1/(s – a)², s > a |
| e^(at) sin(ωt) | ω/[(s – a)² + ω²], s > a |
| e^(at) cos(ωt) | (s – a)/[(s – a)² + ω²], s > a |
These pairs, combined with the derivative formulas, allow efficient solution of many linear ODEs without finding integrating factors or solving characteristic equations in the time domain.
这些变换对与导数公式相结合，使得许多线性常微分方程的求解更加高效，无需在时域中寻找积分因子或求解特征方程。
10. A Worked Example with Polynomial and Trigonometric Terms | 含多项式和三角项的完整例题
Consider the initial value problem: y” – 3y’ + 2y = 4e^(3t), y(0) = 0, y'(0) = 0.
考虑初值问题：y” – 3y’ + 2y = 4e^(3t)，y(0) = 0，y'(0) = 0。
Step 1: Take the Laplace transform of both sides. Using the second-derivative and first-derivative formulas:
第一步：对方程两边取拉普拉斯变换。利用二阶导数和一阶导数公式：
(s²Y – s·0 – 0) – 3(sY – 0) + 2Y = 4/(s – 3)
Step 2: Simplify to obtain (s² – 3s + 2)Y(s) = 4/(s – 3). Note that s² – 3s + 2 = (s – 1)(s – 2).
第二步：化简得到 (s² – 3s + 2)Y(s) = 4/(s – 3)。注意 s² – 3s + 2 = (s – 1)(s – 2)。
Step 3: Solve for Y(s): Y(s) = 4/[(s – 3)(s – 1)(s – 2)].
第三步：解出 Y(s)：Y(s) = 4/[(s – 3)(s – 1)(s – 2)]。
Step 4: Partial fractions: 4/[(s – 3)(s – 1)(s – 2)] = A/(s – 3) + B/(s – 1) + C/(s – 2). Solving gives A = 2, B = 2, C = -4. Therefore Y(s) = 2/(s – 3) + 2/(s – 1) – 4/(s – 2).
第四步：部分分式分解：4/[(s – 3)(s – 1)(s – 2)] = A/(s – 3) + B/(s – 1) + C/(s – 2)。求解得 A = 2，B = 2，C = -4。因此 Y(s) = 2/(s – 3) + 2/(s – 1) – 4/(s – 2)。
Step 5: Take the inverse transform: y(t) = 2e^(3t) + 2e^(t) – 4e^(2t). This satisfies both initial conditions, as y(0) = 2 + 2 – 4 = 0 and y'(0) = 6 + 2 – 8 = 0.
第五步：取逆变换：y(t) = 2e^(3t) + 2e^(t) – 4e^(2t)。该解满足两个初始条件，因为 y(0) = 2 + 2 – 4 = 0 且 y'(0) = 6 + 2 – 8 = 0。
This example demonstrates the entire workflow: transform the differential equation, use derivative formulas, algebraically solve for Y(s), decompose into partial fractions, and invert to find y(t).
该例子展示了完整流程：对微分方程进行变换，使用导数公式，代数求解 Y(s)，分解为部分分式，然后逆变换得到 y(t)。
11. Justification of the General Formula by Induction | 用归纳法证明一般公式
We can rigorously justify the n-th derivative formula using mathematical induction. The base case n = 1 is the first-derivative formula already proven. Assume the formula holds for some k ≥ 1:
我们可以使用数学归纳法严格证明 n 阶导数公式。基础情形 n = 1 即已经证明的一阶导数公式。假设公式对某个 k ≥ 1 成立：
L{f⁽ᵏ⁾(t)} = sᵏ F(s) – sᵏ⁻¹ f(0) – sᵏ⁻² f'(0) – … – f⁽ᵏ⁻¹⁾(0)
Now consider f⁽ᵏ⁺¹⁾ = (f⁽ᵏ⁾)’. Applying the first-derivative formula to g = f⁽ᵏ⁾:
现在考虑 f⁽ᵏ⁺¹⁾ = (f⁽ᵏ⁾)’。将一阶导数公式应用于 g = f⁽ᵏ⁾：
L{f⁽ᵏ⁺¹⁾} = s L{f⁽ᵏ⁾} – f⁽ᵏ⁾(0)
Substituting the induction hypothesis: L{f⁽ᵏ⁺¹⁾} = s[sᵏ F(s) – sᵏ⁻¹ f(0) – … – f⁽ᵏ⁻¹⁾(0)] – f⁽ᵏ⁾(0) = sᵏ⁺¹ F(s) – sᵏ f(0) – sᵏ⁻¹ f'(0) – … – s f⁽ᵏ⁻¹⁾(0) – f⁽ᵏ⁾(0) = sᵏ⁺¹ F(s) – Σ_{i=0}^{k} sᵏ⁻ⁱ f⁽ⁱ⁾(0). This completes the induction.
将归纳假设代入：L{f⁽ᵏ⁺¹⁾} = s[sᵏ F(s) – sᵏ⁻¹ f(0) – … – f⁽ᵏ⁻¹⁾(0)] – f⁽ᵏ⁾(0) = sᵏ⁺¹ F(s) – sᵏ f(0) – sᵏ⁻¹ f'(0) – … – s f⁽ᵏ⁻¹⁾(0) – f⁽ᵏ⁾(0) = sᵏ⁺¹ F(s) – Σ_{i=0}^{k} sᵏ⁻ⁱ f⁽ⁱ⁾(0)。归纳完成。
This proof shows that the formula is not a mere coincidence but follows logically from the fundamental definition and repeated integration by parts.
这个证明表明该公式并非偶然巧合，而是从基本定义和反复分部积分中逻辑推导出来的。
12. Final Summary and Exam Tips | 总结与考试建议
To master the Laplace transform of derivatives, remember these core ideas: the transform of f’ is sF(s) – f(0); the transform of f” is s²F(s) – s f(0) – f'(0); and for higher orders, continue the pattern with descending powers of s and all initial conditions subtracted.
要掌握函数导数的拉普拉斯变换，请记住这些核心理念：f’ 的变换是 sF(s) – f(0)；f” 的变换是 s²F(s) – s f(0) – f'(0)；对于更高阶，继续按照 s 的降幂模式，并减去所有初始条件。
In IB examinations, you may be asked to derive the first-derivative formula, compute transforms of derivatives using known pairs, or solve initial value problems. Be methodical: first transform each term, then solve algebraically, then invert.
在 IB 考试中，您可能会被要求推导一阶导数公式、使用已知变换对计算导数的变换，或求解初值问题。请有条理地操作：先变换每一项，然后代数求解，最后逆变换。
Always verify whether the initial conditions are given at t = 0. Some problems may define y(t) for t < 0 as zero, which aligns with the Laplace transform's one-sided nature. In such cases, the initial conditions at t = 0⁺ are the relevant ones.
始终检查初始条件是否在 t = 0 处给出。有些问题可能将 t < 0 时的 y(t) 定义为零，这与拉普拉斯变换的单边性质一致。在这种情况下，t = 0⁺ 处的初始条件才是相关的。
Practice with a variety of functions: polynomials, exponentials, sinusoids, and their products. Each time you apply the derivative formula, check that the resulting expression matches direct computation when possible. This reinforces both accuracy and conceptual understanding.
请使用多种函数进行练习：多项式、指数函数、正弦
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