I am trying to make an introduction and make myself comfortable with the Nonstandard Analysis in order to gain intuition for derivatives and integration. I am trying to prove myself the famous Chain Rule for differentiation by using the principles of Nonstandard Analysis but I am stuck at somewhere and I am trying to understand where I am doing wrong.
Here is the procedure I am following.
We have two functions $x=f(t)$ and $y=g(x)$ hence $y=g(f(t))$ $f$ has derivative at $t$ and $g$ has derivative at $f(t)$.
As it is widely known, chain rule states that the derivate of $y$ with regards to $t$ is:
$\frac{dy}{dt} = g'(f(t))f'(t) = \frac{dy}{dx}\frac{dx}{dt}$
Now, I first write the slope equation of the $x=f(t)$ with regards to $t$:
$\frac{\Delta{x}}{\Delta{t}} = f'(t) + \epsilon$
Whre $\epsilon$ is an infitesimal. Now according to Increment Theorem, it is
$f(t + dt) - f(t) = \Delta{x} = f'(t)dt + \epsilon dt$
Where $dt = \Delta{t}$. So we can write the differential of x as $dx = f'(t)dt$.
Now I move to the $y=g(x)$ and apply the same methodology. The independent variable $t$ has increased by $dt$ which causes an infitesimal change $\Delta{x}$ in the dependent variable $x$, so the slope equation for $g(x) = g(f(t))$ at $x = f(t)$ is:
$\frac{\Delta{y}}{\Delta{x}} = g'(f(t)) + \delta$
Where $\delta$ is infinitesimal.
Applying Increment Theorem:
$g(f(t + dt)) - g(f(t)) = g(x + \Delta{x}) - g(x) = \Delta{y} = g'(f(t))\Delta{x} + \delta \Delta{x}$
Now I divide all sides of the above equation by $dt$ which is equal to $\Delta{t}$:
$\frac{\Delta{y}}{dt} = g'(f(t))\frac{\Delta{x}}{dt} + \delta \frac{\Delta{x}}{dt}$.
Taking the standard part of the above expression should yield the derivative we are looking for:
$st(\frac{\Delta{y}}{dt}) = st(g'(f(t))\frac{\Delta{x}}{dt}) + st(\delta \frac{\Delta{x}}{dt}) = g'(f(t))st(\frac{\Delta{x}}{dt}) + 0 = g'(f(t))f'(t)$
The line above actually completes the proof, but what I want is to express the equation in the form of $\frac{dy}{dx}\frac{dx}{dt}$ as well, since it is not a mere notation anymore in Nonstandard Analysis and all $dy, dx, dt$ have actual values attached to them. This where I run into a problem.
We know that it is $\frac{dx}{dt} = f'(t)$ already. But from the Increment theorem of $\Delta{y}$ it is $\Delta{y} = g'(f(t))\Delta{x} + \delta \Delta{x}$, so, the differential $dy$ is:
$dy = g'(f(t))\Delta{x}$ and $\frac{dy}{\Delta{x}}=g'(f(t))$.
Now I substitute both $g'(f(t))$ and $f'(t)$ with differential representations and it becomes:
$g'(f(t))f'(t) = \frac{dy}{\Delta{x}} \frac{dx}{dt}$
As you can see $\Delta{x}$ and $dx$ are different values and do not cancel each other out. So I cannot write the chain rule in the form of differentials. I think I am making a mistake in the procedure I am following but I cannot figure it out, so I turned here for help. What is incorrect here?
Thanks in advance.
Instead of introducing $dx$ as $f'(x)dx$, I would work with $\Delta t$ and $\Delta x$ throughout. This way one gets the relation $\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x} \frac{\Delta x}{\Delta t}$. Note that there is a technical point here that needs to be handled when $\Delta x$ vanishes (in which case one can't divide by it). This is a relation among hyperreal quantities rather than real quantities, and is therefore not the chain rule yet. It is only at this stage that I would apply the standard part function to the relation $\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x} \frac{\Delta x}{\Delta t}$, so as to get the relation between the derivatives.