Suppose we have the function
$$M(t)=ce^{rt}$$
where $c$ is a constant.
Trivially, it follows that $\frac{dM_t}{dt}=r\cdot c e^{rt}=rM_t$, hence we can write this in the form
$$\frac{dM_t}{M_t}=rdt$$
I'm trying to derive the same equation by looking at the change $dM_t$
I have $$dM_t=M_{t+dt}-M_t=ce^{r(t+dt)}-ce^{rt}=M_t(e^{rdt}-1)$$
Hence, I'd expect to have
$$e^{rdt}-1=rdt$$
which makes no sense to me.
As far as I know, the derivative with respect to a variable is the change when all the other variables are held fixed. Can someone explain to me where's the issue and in general, how to relate the change to the derivative especially in the case when the function depends on more than $1$ variable?
If ${\rm d}t$ is small then
$$ e^{r{\rm d}t} = 1 + \frac{(r{\rm d}t)}{1!} + \frac{(r{\rm d}t)^2}{2!} + \cdots = 1 + r{\rm d}t $$
So you recover your expression. As for the change of a function when it depends on more than one variable, let $f = f(x_1, x_2, \cdots)$
$$ {\rm d}f = \sum_k\frac{\partial f}{\partial x_k}{\rm d}x_k = \nabla f\cdot {\rm d}{\bf x} $$