Derivative of the Laplace Transform

Question

I calculated that $$ \mathcal{L}'[f(t)] = \mathcal{L}[f'(t)]-s\mathcal{L}[f(t)] $$ And if I try to substitute some values for f(t) I always get zero. So, how do I demonstrate that it gives, or doesn't always give, zero?

Answer 1

It's important to note that the Laplace transform is a linear operator, which means it satisfies the following properties for any functions $f(t)$ and $g(t)$:

$\mathcal{L}[af(t) + bg(t)] = a\mathcal{L}[f(t)] + b\mathcal{L}[g(t)]$ (where $a$ and $b$ are constants) $\mathcal{L}[f'(t)] = s\mathcal{L}[f(t)] - f(0)$ Using these properties, we can simplify the expression for $\mathcal{L}'[f(t)]$ as follows:

\begin{align*} \mathcal{L}'[f(t)] &= \mathcal{L}[f'(t)] - s\mathcal{L}[f(t)] \ &= (s\mathcal{L}[f(t)] - f(0)) - s\mathcal{L}[f(t)] \quad\quad \text{(using property 2)}\ &= -f(0) \end{align*}

Therefore, we can see that $\mathcal{L}'[f(t)]$ is not always zero. It depends on the value of $f(0)$.

For example, if $f(t) = \sin(t)$, then $f(0) = 0$ and $\mathcal{L}'[f(t)] = 0$. However, if $f(t) = \cos(t)$, then $f(0) = 1$ and $\mathcal{L}'[f(t)] = -1$.