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Below is a quote from Options, Futures, and Other Derivatives (9th Edition) by John C. Hull p.319, where $G$ is a continuous and differentiable function of two variables $x$ and $y$, $\Delta x$ is small change in $x$ and $\Delta G$ is the resulting small change in $G$:

the Taylor series expansion of $\Delta G$ is $$ \Delta G=\frac{\partial G}{\partial x} \Delta x+\frac{\partial G}{\partial y} \Delta y+\frac{1}{2} \frac{\partial^2 G}{\partial x^2} \Delta x^2+\frac{\partial^2 G}{\partial x \partial y} \Delta x \Delta y+\frac{1}{2} \frac{\partial^2 G}{\partial y^2} \Delta y^2+\cdots \tag{14A.3} $$ In the limit, as $\Delta x$ and $\Delta y$ tend to zero, equation (14A.3) becomes $$ d G=\frac{\partial G}{\partial x} d x+\frac{\partial G}{\partial y} d y \tag{14A.4} $$

I see that (14A.3) follows a two-variable Taylor series expansion of $G(x+\Delta x,y+\Delta y)$ at $(x,y)$, but I cannot follow how (14A.3) becomes (14A.4), which looks like a total derivative of $G$, in the limit as $\Delta x$ and $\Delta y$ tend to zero. How is (14A.4) derived from (14A.3) by taking the limit?

icedcoffee
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    You don’t derive this from the Taylor expansion. This is a very confusing way of saying things, as they’re mixing up definitions and theorems (and seriously, there’s no need to bring in Taylor at this stage). And if we’re being literal and serious, the limit as $\Delta x,\Delta y\to 0$ is $0$, so you end up with $0=0$. See here (for 2D) and here (for 1D) for how to interpret the equality (essentially an explanation of why the definition of derivative is the way it is). – peek-a-boo Aug 06 '23 at 13:08

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There's a rigorous answer using analysis but given this doesn't come from an analysis text, I will omit the overly sophisticated answer.

Intuitively speaking:

In the limit case as $\Delta x \rightarrow 0, \Delta y \rightarrow 0$, terms like $$(\Delta x)^2, (\Delta y)^2, \Delta x\Delta y$$ become negligible, and the same goes for any terms of higher order. Loosely speaking, any non-linear term i.e. any term that is not $$\frac{\partial G}{\partial x}\Delta x, \frac{\partial G}{\partial y}\Delta y$$

Will be truely considered zero / negligible. This is because as we tend to $\Delta x \rightarrow 0, \Delta y \rightarrow 0$, the linear terms dominate as the most significant.

One way of this intuitively making sense is imagine I had a function of one variable and I found: $$\Delta f = \Delta x + \frac{1}{2}(\Delta x)^2$$

In the limit:

$$df = dx + \frac{1}{2}(dx)^2$$

Now considering integrating this (again omitting any rigor):

$$f = \int (dx + \frac{1}{2}(dx)^2) = x + \frac{1}{2} \int (dx) dx = x$$

The $dx^2$ term is just so much "smaller".

In short that is why you can ignore those higher order terms intuitively. You can also take comfort in the fact there are more rigorous arguments to show so.

PH Herman
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    this isn’t just omitting rigor (which is fine since you explicitly mention it), but you’re confusing the logic of things. One ought to maintain a clear distinction (atleast when first presenting a given topic) between a definition, a theorem, motivation for the definition, and motivation for a theorem. – peek-a-boo Aug 06 '23 at 13:21
  • I argue such pedantics are not particularly necessary for this question, especially since a) the textbook went with the route of "deriving" the total derivative formula from the multivariable taylor series expansion. Which is super goofy, and is drowning in too many assumptions about the differentiability of G. BUT I guess it is a weird way of motivating the definition of the total derivative formula. hATe finance texTbooKs, was just working with the ingredients there. – PH Herman Aug 14 '23 at 11:50