Why doesn't the definition of derivative generalize smoothly from single variable to multivariable calculus like the definition of continuity?

As from my previous questions, one may follow that I am trying to understand the differences between real and complex derivative.

My question is:

For a function $f:\mathbb{R}\to\mathbb{R}$, the derivative at a point $c\in\mathbb{R}$ is defined as $$f'(c)=\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}$$

For a complex function $f:\mathbb{C}\longrightarrow\mathbb{C}$ the complex derivative at a point $z$ is defined as $$f'(z)=\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}.$$

But for a function $f:\mathbb{R}^2\longrightarrow \mathbb{R}^2$, why isn't the derivative at a point $(x,y)$ defined as $$f'(x,y)=\lim_{(h,k)\rightarrow (0,0)}\frac{f((x,y)+(h,k))-f(x,y)}{(h,k)}?$$

Also, in this question, they say, Differences between the complex derivative and the multivariable derivative., that it is because, it is not possible to define division in $\mathbb{R}^2$. But can we not define the inverse of $(h,k)$ as $(\frac{1}{h}, \frac{1}{k})$, for $h\neq 0, k\neq 0$? And define $(a,b){(c,d)}^{-1}$ this way?

$\newcommand{\d}{\mathrm{d}}$The reasons why your definition fails has been discussed appropriately in the comments, but as you asked I will provide an explicit example of the failure of your definition. I will also showcase the true definition, and link it to the derivatives with which you’re familiar.

Counter example:

Let $f(x,y)=(x^2+y^2,x^2-y^2)$ and let's differentiate it at $(0,0)$. Its Jacobian, i.e. its proper derivative (differential) according to the standard definition, is: $$\d f=\begin{pmatrix}2x&2y\\2x&-2y\end{pmatrix}\overset{(0,0)}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$Let's attempt your definition. Consider the numerator of your quotient: $$f((0,0)+(h,k))-f(0,0)=(h^2+k^2,h^2-k^2)$$And now let's "divide" it by $(h,k)$, using your pointwise operation and allowing for dodgy notation: $$\frac{\Delta f}{\Delta(x,y)}=\left(\frac{h^2+k^2}{h},\frac{h^2-k^2}{k}\right)=(h+k^2/h,h^2/k-k)$$What happens as $(h,k)\to(0,0)$? The limit exists if it is the same regardless of the path I take and if it exists for all paths. The path $h=k\to0$ clearly has limit $(0,0)$, but the path $h=k^2$ has: $$\lim_{k\to0}(k^2+1,k^3-k)=(1,0)$$So we see that the limit does not exist, as I have achieved two different answers using two paths to $(0,0)$. Moreover, even if it did exist, in what way would it be a derivative? Your answer would be a vector, but a derivative needs to be a linear map, unless you really want to shake up the definitions...

In the spirit of generalising the derivative as a linear approximation, consider this definition:

If $f:X\to Y$, where $X,Y$ are metric spaces, $f$ is defined to be differentiable at $p_0\in X$ if there exists a unique linear map $\d f:X\to Y$ and a function $\psi:X\to Y$ which is continuous near $p_0$ and $|\psi(p-p_0)|\in o|p-p_0|$, so that: $$f(p)-f(p_0)=\d f\circ(p-p_0)+\psi(p-p_0)$$

If $X,Y=\Bbb R^n,\Bbb R^m$, then $\d f$ would be the Jacobian matrix - remember that linear maps over finite dimensional vector spaces can always be represented by matrices.

If that definition is new to you, I reiterate that the motivation is that this takes the notion of local linear approximation (like the "gradient of a line" in the single variable case) and generalises it easily to fairly arbitrary situations. It keeps the idea that if I zoom in on the graph, or here generally if I zoom in on the surface, there comes a point where it appears “flat”, and the derivative (or more often, the differential), is the linear map that resembles this “flat” section of the surface - recall that matrices can describe planes and other higher dimensional surfaces.

Notice that if $X=Y=\Bbb R$, $\d f$ is a real scalar, let's call it $f'(p_0)$, and we can write:

$$f(p)-f(p_0)=f'(p_0)(p-p_0)+\psi(p-p_0)\implies\frac{f(p)-f(p_0)}{p-p_0}\approx f'(p)$$

Where the approximation is precise as $p\to p_0$, since $|\psi(p-p_0)|\in o|p-p_0|$. Notice that this is precisely the definition of derivative that you use for $f:\Bbb R\to\Bbb R$.

What about the complex derivative?

Well, if we consider $f:\Bbb C\to\Bbb C$ differentiable at $z$ only if it is differentiable as a function of $\Bbb R^2\to\Bbb R^2$, the differential in the sense of $\Bbb R^2$ is a Jacobian matrix with respect to the real and imaginary parts of $f(z)=u+iv$, and the real and imaginary parts of the input $z=x+iy$:

$$f'(z)=f’(x,y)=\begin{pmatrix}\partial_x u&\partial_y u\\\partial_x v&\partial_y v\end{pmatrix}$$

But really this is a complex function, so this matrix must represent a complex scalar. Recall that a complex number $a+ib$ has the matrix representation:

$$a+ib\simeq\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$

Equating the matrices, the derivative $f'=a+ib$ has real part $a=\partial_x u=\partial_y v$, and imaginary part $b=\partial_x v=-\partial_y u$. These are precisely the Cauchy-Riemann equations, which are quite restrictive; $f$'s differentiability as a function of $\Bbb R^2$ does not imply differentiability as a function of $\Bbb C$.

If the notion of matrix representations of $\Bbb C$ is unfamiliar to you, just try writing two complex variables in rectangular form and test adding and multiplying them. Do the same with their matrix representations as above. You should find that they are the same! Then the matrix representation perfectly captures the basic arithmetic operations. Whether the representation holds true in the context of more complicated operations I’m unsure of, but that level of representation theory isn’t relevant here. Things like the matrix exponential and analytic matrix functions do exist however...

Hopefully that explains the multivariable derivative's link to single variable derivatives.