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.

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    $\begingroup$ When $x$ is treated as the independent variable in calculating the derivative of $y=g(x)$, by definition $dx$ is equal to $\Delta x$. You are getting needlessly involved with the differential notation. By definition $g'(x)$ is the standard part of $\frac{\Delta y}{\Delta x}$. $\endgroup$ – Mikhail Katz Sep 30 '13 at 8:54
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    $\begingroup$ If you prefer to use the differential notation throughout, what you could say is that once $dx$ is chosen in such a way that $dx=f'(x)dt$, we choose the infinitesimal increment $\Delta x$ of the independent variable $x$ to be equal to $dx$ when we deal with $y=g(x)$. If you are bothered by the clash of notation with the previous $\Delta x$ (when $x$ was the dependent variable of the relation $x=f(t)$), you can just denote the previous one by $\Delta' x$ instead, since in any case it is not present in the final answer $dy/dt = dy/dx\;dx/dt$. $\endgroup$ – Mikhail Katz Sep 30 '13 at 9:10
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    $\begingroup$ The differential approach does work and is compatible with the chaining behavior, since it produces the relation $\frac{\Delta y}{\Delta t}= \frac{\Delta y}{\Delta x}\frac{\Delta x}{\Delta t}$. The advantage of working with infinitesimals here is that one has the tool of the standard part. Namely, applying the standard part to this relation to obtain the chain rule can only be implemented when there are infinitesimals around. It cannot be implemented in the context of the real continuum alone... $\endgroup$ – Mikhail Katz Oct 1 '13 at 7:22
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    $\begingroup$ ... Your problem seems to be that you want to eat the cake and have it, too: namely to preserve the chaining behavior and also have the ratio of infinitesimals equal the derivative on the nose. $\endgroup$ – Mikhail Katz Oct 1 '13 at 7:22
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    $\begingroup$ I agree with your comments. In the formula $\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}$ the two symbol $dx$ do not have the same meaning, so you cannot literally cancel them. Nonetheless, what I was trying to point out that it is still possible literally to cancel out the infinitesimal in the numerator and denominator in the formula $\frac{\Delta{y}}{\Delta{t}} = \frac{\Delta{y}}{\Delta{x}}\frac{\Delta{x}}{\Delta{t}}$, which shows the advantage of the infinitesimal method. Namely, the advantage is that this formula can be immediately related to chain rule $g(f(x))'=g'(f(x))f'(x)$... $\endgroup$ – Mikhail Katz Oct 1 '13 at 11:22

There's a much easier way to do this. You don't even need the increment theorem for it.

First, let's limit our focus to functions defined at the point $(0, 0)$. If we can prove the chain rule for $f_0$ and $g_0$ at $(0, 0)$, then we can prove it for any two functions $f$ and $g$ at any point $(t, y)$ by setting $f_{0}=a\mapsto f\left(t+a\right)-f\left(t\right)$ and $g_{0}=a\mapsto g\left(f\left(t\right)+a\right)-g\left(f\left(t\right)\right)$.

Second, let's think of functions as constrained sets of points. The antecedent of the chain rule postulates that there are two real numbers $\dot x, \dot y$, and a set of infinitessimal triples:

$\left(\Delta t, \Delta x, \Delta y\right)$

Constrained by:

  • $\Delta t \neq 0$

  • $\exists \epsilon_x\ \left(\Delta x = \left( \dot x+\epsilon_x \right)\Delta t\right)$

  • $\exists \epsilon_y\ \left(\Delta y = \left( \dot y+\epsilon_y \right)\Delta x\right)$

(In other words: $\Delta y$ is approximately proportional to $\Delta x$, which in turn is approximately proportional to $\Delta t$. $\dot y$ and $\dot x$ are the respective proportionality constants.)

And the consequent claims that $\Delta t$ and $\Delta y$ are constrained by:

$\exists \epsilon_{xy}\ \left(\Delta y = \left( \dot x \dot y + \epsilon_{xy} \right)\Delta t\right)$

Assuming the antecedent, the consequent can be proven using substitution:

$\Delta y = \left(\dot y + \epsilon_y\right)\Delta x = \left(\dot y + \epsilon_y\right)\left(\dot x + \epsilon_x\right)\Delta t = \left( \dot x\dot y + \epsilon_x\dot y + \epsilon_y\dot x + \epsilon_x \epsilon_y \right) \Delta t$

The existence of $\epsilon_{xy}$ is thus proven by constructing it as $\epsilon_x\dot y + \epsilon_y\dot x + \epsilon_x \epsilon_y$.

Differentials don't really have a place in this. You don't need them. I think I see what you want them for: you want real numbers that characterize the relationships between $t$ and $x$; $x$ and $y$; $t$ and $y$. And you want to call these numbers $\frac{dx}{dt}$, $\frac{dy}{dx}$, $\frac{dy}{dt}$.

The thing is, there are already numbers for this purpose: $\dot x$, $\dot y$, $\dot x \dot y$. (aka $f^\prime(t)$, $g^\prime(f(t))$, $g^\prime(f(t))f^\prime(t)$.)

Furthermore, if $dt$, $dx$, $dy$ are just values, then $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ is just a tautology. In order to express anything at all using that equation, you have to equivocate between $dy = \dot x \dot y dt$ and $dy = \dot y dx$, and you have to equivocate between $dx = \Delta x$ and $dx = \dot x dt$.

Differentials work when you only want to refer to one variable's dependence on one other variable.


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