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I want to extend the concept of derivative of a real function of real variable to a function $f:A\subset \mathbb{R}^n \to \mathbb{R}^m$ with $A$ open. If $x_0 \in A$ then I say that $f$ has derivative $f'(x_0) \in \operatorname{Hom}(\mathbb{R}^n, \mathbb{R}^m)$ if $$ \lim_{h\to 0} \frac{|f(x_0+h)-f(x_0)-f'(x_0)(h)|}{|h|}=0 .$$ When the function is given in explicit algebraic form as $f(x_1,\dots,x_n) = \sum \limits_{i=1}^m f_{i}(x_1,\dots,x_n)e_i$ and I know that the derivative exists at $x_0\in A$, then I can compute $f'(x_0)$ explicitly because $$ [f'(x_0)]_{ij} = (D_jf_i)(x_0) $$ is the matrix representation of $f'(x_0)$ relative to standard bases and I know from basic calculus how to compute those partial derivatives.

My question is: if I can compute partial derivatives in $x_0$ without knowing if $f'(x_0)$ exists, is there some regularity condition on partial derivatives that is equivalent to the existence of $f'(x_0)$? The existence of partial derivatives isn't sufficient (for $n>1$), for example $\frac{xy}{x^2+y^2}$ has partial derivatives but isn't continuous in $(0,0)$ if is defined $0$ in $(0,0)$ and thus can't be differentiable there. Coming back to the general problem, if partial derivatives exist and are bounded in a neighborhood of $x_0$, then $f$ is continuous in $x_0$ but I believe it could be not differentiable, although I can't write a counterexample. If the partial derivatives are continuous in $x_0$ then $f'(x)$ should exist in a neighborhood of $x_0$ and should be continuous in $x_0$, but this is clearly more than differentiability in $x_0$.

NEWS Differentiability seems to be a slippery regularity. I discovered from Rudin "Principles of Mathematical Analysis" that an equivalence exists for continuously differentiable functions, in fact $f\in C^1(A)$ if and only if $D_jf_i\in C^1(A)$ for all $i,j$. The question seems difficult because of the fact that the regularity of a multivariable function and that of its partial derivatives seem to have a weak connection, although a general heuristic principle for this type of problems could be: a stronger regularity on partial derivatives implies a weaker regularity on the function.

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Just to confirm one of your suspicions: $f$ defined by $f(x, y) = \frac{x^3}{x^2 + y^2}$ for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$ is continuous at $(0, 0)$, but not differentiable there (I learned this from Rudin's book). –  Dylan Moreland Jul 25 '11 at 16:40
The references in my answer to the question at math.stackexchange.com/questions/51296 may be of use. –  Dave L. Renfro Jul 25 '11 at 17:07

2 Answers 2

I guess you are looking for the below theorem: Reference Mathematical Analysis, by Tom Apostol page $357$.

$\textbf{Theorem.}$ Assume that one of the partial derivatives $D_{1}\mathbf{f},\cdots D_{n}\mathbf{f}$ exists at $\mathbf{c}$ and that the remaining $n-1$ partial derivatives exists in some $n$-ball $B(\mathbf{c})$, and are continuous at $\mathbf{c}$. Then $f$ is differentiable at $\mathbf{c}$.

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This theorem states a sufficient condition for differentiability in $x_{0}$ but I believe it cannot be an equivalence. In fact, consider $f:\mathbb{R}^n\to\mathbb{R}$ defined as $f(v)=|v|^2$ when $|v|\in\mathbb{Q}$ and $f(v)=0$ otherwise, this should be differentiable in 0 with $f'(0)=0$, $(D_{i}f)(0)=0$ for $i=1,\dots,n$ but partial derivatives don't exist in a neighborhood of $0$ –  Marco Castronovo Jul 27 '11 at 5:40

Good question. The function $\chi_{\mathbb{Q}^2}\cdot\|.\|_{\mathbb{R}^2}$ has derivative $0$ at $0$, but has partial derivatives nowhere else. This means there can be no regularity condition on the partials equivalent to differentiability, so the theorem Chandrasekhar quoted is just sufficient.

By the way,

If the partial derivatives are continuous in $x_0$ then $f'(x)$ should exist in a neighborhood of $x_0$ and should be continuous in $x_0$

is incorrect, only $f'(x_0)$ is guaranteed to exist, although $f'$ is indeed continuous at $x_0$.

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