Say we have a continuous LTI system of which we know the analytical expression of its step response. Let's call it $y_{\text{step}}(t)$. Having $y_{\text{step}}(t)$ (and therefore also $Y_{\text{step}}(\omega)$) how can we find the impulse response $y_{\text{impulse}}(t)$?
I read on this question that "With an LTI system, the impulse response is the derivative of the step response. Because the impulse function is the derivative of the step function. Derivative in, derivative out.", while it makes sense to me intuitively, how would one go about proving this?
Any help would be greatly appreciated!
My effort:
We know that: \begin{equation} \frac{df}{dt} = \lim_{h \rightarrow 0}\frac{f(t+h) - f(t)}{h} \quad \quad \text{and} \quad \delta(t)=\frac{du}{dt}, \end{equation} where $\delta(t)$ and $u(t)$ are the dirac distribution and step function respectively.
Due to our system being linear we know that the additivity and homogeneity properties hold. That is, if $S[\cdot]$ is the output of the system for the input $\cdot$, then: \begin{align*} S[x_1(t) + x_2(t)] &= S[x_1(t)] + S[x_2(t)] &\text{(Additivity)} \\ S[ax(t)] &= aS[x(t)] &\text{(Homogeneity)} \end{align*}
Therefore, using the fact that: \begin{equation} \delta(t)=\frac{du}{dt} = \lim_{h \rightarrow 0}\frac{u(t+h) - u(t)}{h} \end{equation} we write: \begin{align*} S[\delta(t)] &= S\left[\lim_{h \rightarrow 0}\frac{u(t+h) - u(t)}{h} \right] \\[10pt] &= \lim_{h \rightarrow 0} \frac{S[u(t+h)] - S[u(t)]}{h} \\[10pt] &= \frac{d\left( S[u(t)]\right)}{dt} \end{align*}
Is my effort correct? Is it true in general for L.T.I systems that \begin{equation} S\left[\frac{dx}{dt}\right] = \frac{d\left(S[x]\right)}{dt} \quad ? \end{equation}