Differentiability of a multi variable function.

Question

My question is, like in the case of complex differentiation for a function to be differentiable at a point (x,y) the derivative must be same no matter which direction is chosen.

So, if I have a function z=f(x, y) shouldn't the derivative from every direction must be same to make it differentiable. That is, shouldn't the partial derivatives be same because the direction used in those cases are x and y?

Explain for a vector functions too.

Answer 1

Good question. I think the answer is the difference in the target space. In complex case, you function maps into $\mathbb{C}$ while for two-variable case, it maps into $\mathbb{R}$. The derivative in the latter case is a linear map, i.e a matrix. Notice that multiplication by a constant complex value is a map from $\mathbb{C}$ to $\mathbb{C}$ that can alternatively be viewed as a linear map from $\mathbb{R}^2$ to $\mathbb{R}^2$. So, there is some analogy. But the reason for the difference in directional derivatives is due to two distinct definitions of differentiability.

Added: (Definitions)

Definition 1: For a map $f: \mathbb{R}^n \to \mathbb{R}^m$, we say $f$ is differentiable at $x$ if there exists a linear map $L:\mathbb{R}^n \to \mathbb{R}^m$ such that $\forall v \in \mathbb{R}^n$ the following limit exists $$ \lim_{\|v\|\to 0}\frac{\|f(x+v)-f(x)-L.v\|}{\|v\|} = 0 \ .$$ We call $L$ the derivative of $f$ at $x$, denoted by $Df(x)$. The value $Df(x).v$ is the directional derivative of $f$ at $x$ in direction $v$.

In case $\mathbb{R}^2 \to \mathbb{R}^2$ the derivative will be a linear map $\mathbb{R}^2 \to \mathbb{R}^2$, i.e. a $2 \times 2$ matrix.

Definition 2: For a map $f:\mathbb{C} \to \mathbb{C}$, we say that $f$ is (analytic) differentiable at $z$ if there exists a single $z_0 \in \mathbb{C}$ such that $$ \lim_{w \to 0} \frac{f(z+w)-f(z)}{w} = z_0 \ . $$ Now, if we identify $\mathbb{C} = \mathbb{R}^2$ and look at $f$ as $\mathbb{R}^2 \to \mathbb{R}^2$ and rewrite the fraction above as directional derivatives, then being analytic differentiable is saying that $f$ is differentiable in the sense of the definition above, BUT, such that the derivative, the $2 \times 2$ matrix, has a special form. The special form guarantees that the matrix action here is equivalent to multiplication by a complex number. The condition for a matrix to be of this type is exactly the criteria on the partial derivatives that make complex valued functions analytic differentiable. Oh, did I forget to say that in definition 1, the derivative matrix is just the matrix of partial derivatives?! You must write down a concrete example to "see" things.

Answer 2

Why you say that derivative must be the same if taken in different directions? Consider that, for example, in real life, you can go to some place at certain velocity but if you take another path it may happen that you can go at a different velocity, say when you reach the top of a mountain. Don't forget that math is cheap physics, i.e., its connected with reality.