1.1.4. Proof of First Derivative

Let f(t) be a function such that f′ (t) exists and is continuous on [0, ∞). Then: L{f′ (t)}=sF(s)−f(0)

This chunk introduces the Laplace Transform of the first derivative of a function. The notation L{f′(t)} denotes the Laplace Transform of the derivative of f(t). According to the formula, the Laplace Transform of f′(t) is equal to s times the Laplace Transform of f(t), denoted as F(s), minus the value of the function f(t) at t=0, which is f(0). This means that when taking the Laplace Transform of the derivative, we not only account for how the function changes (the derivative) but we also need to adjust by the original function's value at the starting point.

Think of the function f(t) as the position of a car over time. Its derivative, f′(t), is the speed of the car. The Laplace Transform of f′(t) helps us analyze the speed in a transformed space where we can solve problems more easily, similar to how knowing both the speed and the initial position of the car helps us understand its journey.

Proof: Using integration by parts: L{f′ (t)}=∫e−stf′ (t)dt=[e−stf(t)] +s∫e−stf(t)dt

This chunk describes the proof process for the Laplace Transform of the first derivative using integration by parts. Integration by parts is a technique from calculus that allows us to rewrite integrals in a different form. Here we start with the integral of e raised to a power times f′(t). After applying integration by parts, we obtain two parts: the boundary term [e−stf(t)] evaluated from 0 to infinity and another integral involving f(t). The boundary term approaches zero as t approaches infinity, so we can simplify the expression to arrive at L{f′(t)} = sF(s) − f(0).

Imagine you're tracking the growth of a plant over time. If you know its height at various moments (the integrals) and how much it grows at those moments (the derivative), using integration by parts allows you to get a complete picture of how height relates to growth over time, ultimately simplifying the process of understanding the plant's overall height at the end.

As t→∞, e−stf(t)→0 (since f(t) is of exponential order). So: L{f′ (t)}=sF(s)−f(0)

This chunk concludes the proof by discussing the behavior of the term e−stf(t) as time approaches infinity. Since f(t) is of exponential order, the product e−stf(t) goes to zero when we take the limit as t approaches infinity. This is crucial because it allows us to disregard this term when calculating the Laplace Transform, leading us to the final result, confirming the relationship found earlier: L{f'(t)} = sF(s) - f(0).

Consider a balloon that is slowly deflating as time passes. Initially, it might have a certain volume (similar to f(0)), but as time goes on, and the balloon gradually deflates, its volume approaches zero. Just as we can ignore the negligible volume of a fully deflated balloon when calculating its final state, we disregard the term e−stf(t) in our calculations as t grows very large, simplifying our analysis.

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