what would happen if directly expand the limit expression of derivative to multivariable function? [I've seen a similar question "https://math.stackexchange.com/questions/2389074/multivariable-derivative-limit-definition", but it doesn't give me a satisfied answer so I choose to post a new one]

I know the limit expression of derivative of a single variable function $y=f(x)$ at $x=c$ is
$$f'(c)=\lim\limits_{\Delta c \to 0} \frac{\Delta f(c)}{\Delta c}=\lim\limits_{h \to 0} \frac{f(c+h)-f(c)}{(c+h)-c}$$
Now consider a multivariable function $y=f(x_1,x_2,...,x_n)$ defined in a certain sphere of nth-dimensional space with the center $(a_1,a_2,...,a_n)$. The distance $\rho$ between a certain point $(c_1,c_2,...,c_n)$ in the sphere and the center of sphere is
$$\rho=\displaystyle\sqrt{\sum_{k=1}^n(a_k-c_k)^2}$$
If I simply copy the limit expression of derivative to the multivariable case I get
$$\lim\limits_{\Delta \rho \to 0} \frac{\Delta f(c_1,c_2,...,c_n)}{\Delta \rho}=\ ?$$
This limit is certainly not partial derivative from my knowledge (I don't even know if it exists or not), so what does it actual represent if it can exist ?

[Edit]: I realized that $\rho$ and $f(x)$ can form a 2-dimensional plane, what I'm wondering more precisely is that if there is any connection between this limit and partial derivatives ?
 A: You've got a good idea, but it's not quite right.  You are thinking of the derivative as a number, but it's better to think of it as a function.  A lot of differential calculus consists of the attempt to approximate nonlinear functions by simpler functions, often linear ones.  Of course, we can't hope to approximate the sine function, say, everywhere by the same linear function, so we try to approximate locally.
When we say $f(x)-f(a)\approx f'(a)(x-a)$ we are approximating $f(x)$ near $a$ by the estimate $$f(x)\approx f'(a)(x-a)+f(a)$$ and the right-hand side is a linear function.  Of course, if we know what $a$ and $f'(a)$ are, we know what the function is, but in this context, we need to concentrate on the function.
For a function $f:\mathbb{R}^n\to\mathbb{R}$ we want to do the same kind of thing.  Now, we need a linear function of $n$ variables to approximate $f$.  We'll look for a linear transformation.  (If you aren't familiar with this term, it's just a linear polynomial in $n$ variables, with no constant term.) $$A(x_1,\dots,x_n)=a_1x_1+\cdots+a_nx_n$$
So $$f(x+h)-f(x)\approx f'(x)(x-h)$$  becomes $$f(X+h)-f(X)\approx A(X-h)$$ where I've changed to $X$ to emphasize that it's a vector.   (h is also an n-dimensional vector, of course.) Now as you said, the way to express this is that the distance between the two values is close to $0$.
$$\|f(X+h)-f(X)-A(X-h)\|\approx 0$$
Close to $0$ compared to what?  Close compared to the distance between $h$ and the origin.  This leads to the actual definition.  The derivative of $f$ at $X$, if it exists, is the linear transformation $A$ such that $$\lim_{h\to0}\frac{\|f(X+h)-f(X)-A(X-h)\|}{\|h\|}=0$$
One can show that if such an $A$ exists, it is unique.
It turns out that the coefficients of $A$ are just the $n$ partial derivatives of $f$ at $X$.  Suppose $A=Df(a)$ is the derivative of $f$ at some point $a=(a_1,a_2,\dots,a_n)$  Then $$A(x_1,x_2,\dots x_n)=f_1(a)(x_1-a_1)+f_2(a)(x_2-a_2)+\cdots f_n(a)(x_n-a_n)$$
where $f_k(a)=\frac{\partial f}{\partial x_k}\big\lvert_{X=a}$.
You may see $A$ called the "total derivative" or the "differential."
The same idea extends readily to case where $f$ is a vector-valued function of one or more variables.  Basically, we do the same thing in each coordinate.
