# Does the Chain Rule in Leibniz notation cancel terms?

I understand this notation is now a differential operator and this is the limit of a quotient, but Leibniz regarded $$\frac {dy}{dx}$$ as a quotient. In Leibniz's theory where $$\frac {dy}{dx}$$ is a quotient, do the terms in chain rule cancel out?

For instance below, there are two instances of $$du$$. $$y = f(u), u = u_1 = u_2 = g(x)$$ $$\frac {dy}{du_2} \Big|_{g(x)} \cdot \frac {du_1}{dx} \Big|_{x} = (f \circ g)'(x) = (f' \circ g)(x) \cdot g'(x)$$

It's my understanding that: $$(du_1 = g(x+h) - g(x)) \ne (du_2 = (g(x) +h) - g(x))$$ becomes the following under the infinitesimal theory Leibniz used: $$\lim_{h \rightarrow 0} {(du_1 = g(x+h) - g(x)) = (du_2 = (g(x) +h) - g(x))}$$ $$du_1 =du_2$$

and the above terms $$du_1, du_2$$ cancel out, leaving $$\frac {dy}{dx}$$.

If these terms do not cancel and are unequal, why is $$du_1 = du_2 = du$$ used in the definition of the chain rule in modern theory? Are there advantages to viewing things like this from infinitesimal theory like Leibniz?

• what are u_1 and u_2? Commented Feb 27, 2021 at 20:37
• How many times did you already ask a question regarding Leibniz notation :)? Commented Feb 27, 2021 at 20:39
• This is a different question
– Nick
Commented Feb 27, 2021 at 20:40
– Nick
Commented Feb 27, 2021 at 20:41
• @Nick. Of course not. Why would you get that? You get $\frac{f(g(x+h))-f(g(x))}{(x+h)-h}$, If $g(x+h)=g(x)+h$ (using continuity), then you can rewrite as $\frac{f(g(x)+h)-f(g(x))}{h} = \frac{f(g(x)+h)-g(x)}{g(x)+h-g(x)}\cdot\frac{g(x+h)-g(x)}{h}$. Commented Apr 3, 2021 at 2:44

There is no equality using the widely used theory of limits because $$\frac {dy}{dx}$$ is the limit of a quotient, however with Leibniz's theory of infinitesimals $$\frac {dy}{dx}$$ is a quotient. Leibniz's infinitesimal theory has logical contradictions and to resolve them the theory of non-standard analysis on the hyperreal numebers can be used. The hyperreals $$\mathbb{R}*$$ are an extension to the reals with infinitesimals $$\mathbb{R} \subseteq \mathbb{R}^*$$. The derivative and integral of calculus are expressed with the standard part function which is $$s : \mathbb{R}^* \rightarrow \mathbb{R}$$.
With non-standard analysis the above equality $$du_1 = du_2 = g(x+h) = g(x) + h$$ holds if it is proven. For instance $$g(x)$$ is continuous because it is differentiable. When proven $$du$$ cancels out. When terms are cancelled out due to equality the following is the result:
$$y = f(x), u = g(x)$$ $$\frac {dy}{du} \frac {du}{dx} = \frac {f(g(x+h)) - f(g(x)))}{g(x+h)-g(x)} \frac {g(x+h) -g(x)}{h} = \frac {f(g(x+h) -g(x))}{h}$$
due to continuity, $$g(x) + h - g(x) = g(x+h) - g(x)$$ then the result is: $$\frac {f(g(x) + h) - f(g(x))} {h}$$
This appears to agree with the result using limits: $$f'(g(x)) = \lim_{h\rightarrow \infty}{\frac {f(g(x) + h) - f(g(x))} {h}}$$