I'd like to compare and contrast the derivative across different areas of mathematics just to organize ideas in my head. In this question, $f(x,y)$ means $f: \mathbb{R}^2 \to \mathbb{R}$ and $f(z)$ means $f: \mathbb{C}\to\mathbb{C}$


In multivariate calculus ($f(x,y)$), in order to find the limit of $f(x,y)$ as $(x,y) \rightarrow (x_0,y_0)$, you need to approach $(x_0, y_0)$ from all possible directions in the $xy$ plane. Likewise in complex analysis, in order to find the limit of $f(z)$ as $z \rightarrow z_0$, you need to approach $z_0$ along every path and check the limit.


In multivariate calculus, you usually talk about a derivative along a single direction. For instance, the partial derivative with respect to $x$ is given by

$$\frac{\partial f}{\partial x} = \lim_{h\to\ 0} \frac{f(x+h,y) - f(x,y)}{h}$$

This is the derivative parallel to the x-axis. You can generalize this to a derivative along any straight line direction. It's called a directional derivative. You can further generalize this to the derivative along any arbitrary path (not just straight lines) in your domain. However the two are sort of the same because at any one point on the arbitrary path, the derivative you compute there is a directional derivative at that instant of the path. Anyways, the point is to show that derivatives are taken with respect to directions.

For complex-valued functions of a complex variable, the derivative is

$$\frac{df(z)}{dz} = \lim_{h\to\ 0} \frac{f(z+h) - f(z)}{h}$$ with the understanding that h is a complex number that approaches $0$ along any path in the complex plane. So the derivative must be checked along every path and every path has to yield the same value of the limit in order for the derivative to exist.


Why isn't there such notion of a derivative for multivariate functions $f(x,y)$? The domain of a function $f(z)$ is the complex plane. It's a plane. The domain of a function $f(x,y)$ is the $xy$ plane. Also a plane. Why can't I ask for the derivative of $f(x,y)$ at a point P where I check all possible paths as $(x,y) \rightarrow P$ just as I did for the derivative of $f(z)$? Why must I always take the derivative in a specific direction? Likewise, why can't you ask for the derivative of $f(z)$ in a certain direction at $z$?

My thoughts

Although a complex variable can take values in a complex plane, it's still a '1 dimensional number.' $z = x + iy$ is usually said to be a '2 dimensional number', but this is from the perspective of real numbers (1D) and generalizing to complex numbers (2D). However if I move the starting point in my brain to complex numbers, then a complex number is just a 1D number. Complex numbers live on a plane, but the plane should be thought of as 1D. Therefore just as checking the derivative of a single variable function $f(x)$ as $h$ goes to $0$ on 'all possible paths' is normal, checking the derivative of $f(z)$ as $h$ goes to $0$ on 'all possible paths' is normal because there aren't any paths (the plane is 1D). I can't ask for the $f'(z)$ along a particular direction because there is 'no direction' just as there is really no direction for a 1D real number line. However, I'm still confused why you can't ask for $f'(x,y)$ without specifying a direction. $f'(x,y)$ is always given with respect to a direction.

  • $\begingroup$ In my opinion, the emphasis on directional derivatives is mostly about introductory texts trying to stay in the realm of one-dimensional calculus as long as possible, not because it's a good way to think about derivatives in higher dimension. $\endgroup$
    – user14972
    Aug 10, 2017 at 2:29
  • 1
    $\begingroup$ First, all directional derivatives exist doesn't mean the function is differentiable. And all the directional derivatives certainly need not be the same even if it's differentiable, so it's unclear what this 'overall' derivative's value would be even if the function were differentiable. Really, if the derivative exists, we have a tangent plane, just as we have a tangent line in 1D and the gradient tells you about the tangent plane, just like the derivative tells you about the tangent line. $\endgroup$ Aug 10, 2017 at 2:32
  • $\begingroup$ My two cents: complex analysis focuses more on holomorphic functions and the important theorems and theories around them. Multivariate calculus focuses more on actually "getting calculations done" even in situations where C-R equations don't hold or when you have more than 2 variables. Same soup from two different viewpoints. $\endgroup$ Aug 10, 2017 at 5:07

3 Answers 3


Derivatives are not defined in terms of approaching all possible paths! For a domain (for now, let's say a domain is a connected open set) $U \subseteq \mathbb{R}^2$ and a function $f : U \to \mathbb{R}$, the derivative of $f$ at a point $P \in U$ is a linear map $Df_P$ from the tangent space of $U$ at $P$, to the tangent space of $\mathbb{R}$ at $f(P)$, which ends up meaning $Df_P : \mathbb{R}^2 \to \mathbb{R}$. It is the unique linear map satisfying $$ \lim_{h \to 0 \in \mathbb{R}^2} \frac{f(P+h) - f(P) - Df_P(h)}{\|h\|} = 0. $$ Bear in mind the $\epsilon-\delta$ definition of limits: this limit is required to converge in the $\epsilon-\delta$ sense. You can prove that it's equivalent to require convergence for every sequence $\{h_n\}$ converging to $0$. But we are not talking about approaching along paths.

The above definition is reasonably strong. Among other things it implies convergence along every path, in particular along straight-line paths: for fixed nonzero $h \in \mathbb{R}^2$, we have $(f(P+th)-f(P) - Df_P(th))/t\|h\| \to 0$. Since $Df_P$ is linear, we have $Df_P(th) = t \, Df_P(h)$ and we can simplify to $\lim_{t \to 0} \frac{f(P+th)-f(P)}{t} = Df_P(h)$. This yields directional derivatives (well, I've presented derivative with respect to a vector; directional derivative is when $\|h\|=1$). The partial derivatives are the directional derivatives for $h = \hat\imath, \hat\jmath$ (the standard basis unit vectors).

The converses do not hold. Partial derivatives can fail to determine directional derivatives in other directions; directional derivatives can exist in every (straight-line) direction but fail to imply the existence of a "total" derivative $Df_P$.

The complex derivative is the same thing plus the requirement that $Df_P$ must be complex linear.


There is one property of complex numbers that $\mathbb{R}^2$ has not, that is the inverse, namely

\begin{equation} z^{-1}=\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2} \end{equation}

This is the property that makes possible to have a derivative from $\mathbb{C} \rightarrow \mathbb{C}$ and not from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$.

You can say that $\mathbb{C}$ is $\mathbb{R}^2$ possessing a structure of a particular product from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$. So the difference with ordinary derivative is not about paths but rather about algebra.

  • $\begingroup$ Why do we assume that if a complex derivative exists, then each for any trajectory, it must be the same? $\endgroup$ Apr 6, 2021 at 19:21

My initial reaction is that the multivariable case is similar to the single real variable case, because you hold the other variables fixed and consider the function as just a function of the one variable with respect to which you are differentiating... You are essentially computing partials. ..

In the complex case, when differentiating, you do involve all directions in the complex plane... It is well known that this is a stronger condition than differentiability wrt a real variable, which results in various nice theorems and simplifications... The first example that comes to mind is the fact that once differentiable (holomorphic ) implies infinitely so in the complex realm, and in fact, analytic (equal to its power series ).

Then, getting back to the multivariable case, you could approach it from the point of view of differential geometry. Say you are working in $\mathbb R^n $, for starters. You get, if I remember correctly, a function $Df $ which consists of matrices full of partial derivatives, which makes the following diagram commute: $$ \require{AMScd}\begin{CD}TM @>Df>>TN\\ @VVV@VVV\\ M@>f>>N \end{CD} $$. Here the downward arrows denote the projections of the tangent bundles of M and N, respectively. ..So, this Df takes all of the variables into account, but still they are built out of partial derivatives. .. (Here M and N are manifolds. Suffice it to say that locally they look like Euclidean space. )


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