First principle of differentiation needs to approximate a sufficiently small integral as area? $$y(t+\Delta t) = e^{- \int_{t}^{t+\Delta t}H(t')dt'}y(t)$$ is the solution to the differential equation $$\frac{dy}{dt} = -H(t)y$$, $H(t)$ and $y$ are scalar.
However, in showing that $$y(t+\Delta t) = e^{- \int_{t}^{t+\Delta t}H(t')dt'}y(t)$$ leads to the differential equation by first principle, something strange happens.
First, I taylor expand $$e^{- \int_{t}^{t+\Delta t}H(t')dt'} = 1 + \left(- \int_{t}^{t+\Delta t}H(t')dt'\right) + \frac{1}{2} \left(- \int_{t}^{t+\Delta t}H(t')dt'\right)^2 +...$$.
Therefore, 
$$ y(t+\Delta t) = y(t) + y(t)\left(- \int_{t}^{t+\Delta t}H(t')dt'\right) + y(t)\frac{1}{2} \left(- \int_{t}^{t+\Delta t}H(t')dt'\right)^2 + ...\\
\lim_{\Delta t \rightarrow 0}\frac{y(t+\Delta t) - y(t)}{\Delta t}=  \lim_{\Delta t \rightarrow 0}\frac{y(t)}{\Delta t}\left(- \int_{t}^{t+\Delta t}H(t')dt'\right) + \frac{y(t)}{\Delta t}\frac{1}{2} \left(- \int_{t}^{t+\Delta t}H(t')dt'\right)^2 + ...$$
However, from that I cannot show that:
$$\lim_{\Delta t \rightarrow 0} \frac{y(t+\Delta t) - y(t)}{\Delta t} = -H(t)y(t)$$
unless $$\lim_{\Delta t \rightarrow 0}\left(- \int_{t}^{t+\Delta t}H(t')dt'\right) = H(t)\Delta t$$. But I do not think that's true and there must be another more rigorous argument. What went wrong here?
 A: This proof uses "Little-o" notation. (read on wikipedia for it's definition and properties)
let's start with $y(t+\Delta t)$, and let's develop it:
$$
y(t+\Delta t)=y(t)+y'(t)\Delta t+o(\Delta t)
$$
now let's tackle this, using the intermediate value theorem for definite integrals:
$$
e^{-\int_{t}^{t+\Delta t}H(t')dt'}y(t)=e^{-H(t+\theta \Delta t)\Delta t}y(t)
$$
where $\theta \in [0,1]$ is chosen appropriately.
Let's develop it now:
$$
e^{-H(t+\theta \Delta t)\Delta t}y(t)=[1-H(t+\theta \Delta t)\Delta t +o(\Delta t)]y(t)=
$$
$$
=y(t)-H(t+\theta \Delta t)y(t) \Delta t +o(\Delta t)y(t)
$$
now let's put equal the two taylor developments:
$$
y(t)+y'(t)\Delta t+o(\Delta t)=y(t)-H(t+\theta \Delta t)y(t) \Delta t +o(\Delta t)y(t)
$$
simplify:
$$
y'(t)\Delta t+o(\Delta t)=-H(t+\theta \Delta t)y(t) \Delta t +o(\Delta t)y(t)
$$
develop the $H$:
$$
y'(t)\Delta t+o(\Delta t)=-H(t)y(t) \Delta t+o(\Delta t) 
$$
divide by $\Delta t$:
$$
y'(t)=-H(t)y(t)+o(1) 
$$
take the limit $\Delta t \to 0$:
$$
y'(t)=-H(t)y(t)
$$
