When are the linear regression parameters of Y and X the same as the parameters of Y' and X'? I am working on some simple linear modeling of a physical system and assumed that taking the derivative of an equation
$$Y = \beta_1 + \beta_2 X + \varepsilon$$
would give me 
$$\frac{dY}{dt} = \gamma_1 + \gamma_2 \frac{dX}{dt} + \omega$$
with $\gamma_1 \approx \beta_1$ and $\gamma_2 \approx \beta_2$ (but no).  
Which assumptions would need to be true for these parameters to be the same?
 A: By "simple linear modeling of a physical system" I understand the following:


*

*There are two physical quantities $x$ and $y$ that both vary differentiably over time in such a way that $y(t) = \beta_1 + \beta_2 x(t)$ for unknown constants $\beta_1$ and $\beta_2.$  We may write them as $x(t)$ and $y(t)$ when we want to be explicit about the dependence on time $t.$

*You observe $x$ and $y$ at various times $t_1 \lt t_2\lt \cdots\lt t_n.$  These constitute your data, denoted $(X_i,Y_i)$ at time $t_i, i=1,2,\ldots, n.$

*These observations will not exactly agree with the true underlying values. 


*

*Either you can control $x(t),$ resulting in $X_i = x(t_i),$ or else you expect the differences between $X_i$ and $x(t_i)$ to be so small that you can neglect them and proceed as if $X_i = x(t_i).$

*You have elected--at least provisionally--to model the differences $\varepsilon_i = Y_i - y(t_i)$ as a random process.


*You assume the $\varepsilon_i$ are independent and identically distributed (iid) with a common zero-mean distribution.

*You use some procedure that applies to this model to estimate the values of the unknown constants from the data.  For the sake of keeping the analysis short and simple, I will assume you are using ordinary least squares (OLS).  This assumes the common distribution of the $\varepsilon_i$ has a finite variance, say $\sigma^2.$
It is not possible to make sense of $Y^\prime(t) = dY(t)/dt$ in this model, nor should we even try.  (In applying the definition of the derivative we would find ourselves trying to compute the limit of a difference quotient of the form $(\varepsilon(t+dt) - \varepsilon(t))/dt$ as $dt\to 0,$ but unfortunately that difference diverges because the distribution of the numerator is the same for all $dt$ while the denominator goes to $0.$)
Instead, I understand your question to concern estimation of $y^\prime(t) = dy(t)/dt.$  According to your model,
$$y^\prime(t) = \beta_2 x^\prime(t).\tag{*}$$
The issue at hand concerns estimating this with data.  You might attempt to use first differences of the data as an approximation, assuming the $dt_i = t_{i+1}-t_i$ are sufficiently small:
$$\eqalign{
\frac{Y_{i+1} - Y_i}{dt_i} &= \frac{(\beta_1 + \beta_2 x_{i+1} + \varepsilon_{i+1}) - (\beta_1 + \beta_2 x_i+ \varepsilon_i))}{dt_i} \\
&= \beta_2 \frac{x_{i+1}-x_i}{dt_i} + \frac{\varepsilon_{i+1}-\varepsilon_i}{dt_i}.
}$$ 
This is a similar-looking model.  If we define the "derived data" via the first differences as
$$Y^\prime_i = Y_{i+1} - Y_i$$
and
$$x^\prime_i = x_{i+1} - x{i},$$
it can be written (upon multiplying by $dt_i$) as
$$Y^\prime_i = \beta_2 x^\prime_i + \phi_i\tag{**}$$
where
$$\phi_i = \varepsilon_{i+1}-\varepsilon_i.$$
Because $\varepsilon_{i+1}$ is independent of $\varepsilon_i,$
$$\operatorname{Var}(\phi_i) = \operatorname{Var}(\varepsilon_{i+1}-\varepsilon_i) = \operatorname{Var}(\varepsilon_{i+1}) + \operatorname{Var}(\varepsilon_i) = \sigma^2 + \sigma^2 = 2\sigma^2$$
and
$$\operatorname{Cov}(\phi_i, \phi_{i+1}) = \operatorname{Cov}(\varepsilon_{i+1}-\varepsilon_i, \varepsilon_{i+1+1}-\varepsilon_{i+1}) = -\operatorname{Var}(\varepsilon_{i+1}) = -\sigma^2.$$
All other covariances $\operatorname{Cov}(\phi_i, \phi_{i+j+1})$ are zero when $j \gt 1.$
Thus the $\phi_i$ are identically distributed but not independent.  Nevertheless, your assumptions tell you their covariance structure up to the multiplicative constant $\sigma^2.$  This is a Generalized Least Squares setting (without an intercept term).   Solutions are relatively straightforward to obtain and they include estimating $\sigma^2.$  They enjoy most of the usual OLS properties.  In particular, the GLS estimates of $\beta_2$ and $\sigma^2$ are unbiased.  That implies you can use this GLS estimate of $\beta_2$ in model $(**)$ to estimate the $\beta_2$ of model $(*).$
Notice that differentiation completely eliminated $\beta_1$ from the model in $(*)$: it is impossible to estimate $\beta_1$ from the derived data.

For solving the GLS Normal equations, note that the covariance matrix $\Omega$ of the $(\phi_i)$ is $\sigma^2$ times the Cartan matrix for the semisimple Lie Algebra $A_{n-1},$ whence its inverse matrix is
$$(\Omega^{-1})_{ij} = \min((n-i)j, (n-j)i) / n$$
for $1 \le i,j\le n-1.$  Use this formula to compute the inverse quickly to high accuracy.
A: Let us assume the equation in levels corresponds to the true data generation process (DGP). Then, in discrete terms (i.e. $dt = 1$), you have that:
$$ dY = \beta_2 dX + \omega $$
where $\omega = d\epsilon$.
This "true" DGP in differences has no constant/intercept. Therefore, a consistent estimator of such constant has mean zero. This is, $E(\gamma_1) \neq \beta_1$, unless the population parameter $\beta_1$ is actually zero.
Regarding $\gamma_2$, its OLS estimator will be a consistent estimator of $\beta_2$ if $dX$ is exogenous. This is, if $E(\omega|dX)=0$, or by the law of iterated expectations, if $E(\omega  dX)=0$.
When would the latter not be true? For example, if your physical system has feedbacks, or if Y and X are part of an equilibrium system. These mean that $X = f(Y)$. In this case, the equation in differences has an endogenous regressor ($dX$), and is said to suffer from simultaneity bias. 
