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There is a theorem in my textbook which says:

Theorem: consider the function $f:\mathbb{R}^2\to\mathbb{R}$ such that both of its partial derivatives at a point exist and at least one of them is continuous, then $f$ is differentiable at that point.

I want to know whether the theorem is true in general for a function $f:\mathbb{R}^n\to\mathbb{R}$, i.e whether the following is true or not:

Consider the function $f:\mathbb{R}^n\to\mathbb{R}$ such that all its partial derivatives exist at a point and at least one of these partial derivatives is continuous at that point,then $f$ is differentiable at that point.

Michael Hardy
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Ibrahim Islam
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    In order to talk about continuity of the partial derivative, you need at the very least assume that the partial derivative exists in a neighborhood (and not only at the point). So you probably want to change the statement of the theorem a bit. – Bart Michels Apr 19 '20 at 10:26
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    According to the quote here https://math.stackexchange.com/a/53711/42969 you need that $n-1$ partial derivatives are continuous at that point. – Martin R Apr 19 '20 at 10:33
  • @Bart Michels is the statement not correct for $f:\mathbb{R^2}\rightarrow\mathbb{R}$ – Ibrahim Islam Apr 19 '20 at 10:48
  • Do I need to edit the statement for $\mathbb{R^2}$ – Ibrahim Islam Apr 19 '20 at 10:50
  • If you add the hypothesis that the partial derivatives exist and are bounded near the point of interest, your theorem becomes true. – Lázaro Albuquerque Apr 19 '20 at 16:37
  • $$\begin{align} \text{right: } & \mathbb{R}^n \ \text{wrong: } & \mathbb{R^n} \end{align}$$ I edited the question accordingly. – Michael Hardy Apr 19 '20 at 16:47

1 Answers1

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Here is a counterexample.

$\displaystyle f(x,y,z)=\begin{cases} \frac{xy}{x^2+y^2}+c(z)\quad (x,y)\ne(0,0) \\ c(z) \quad (x,y)=(0,0) \end{cases}$

where $c(z)$ is any function with continuous derivative (so that $f_z$ exists).

Since $\frac{xy}{x^2+y^2}$ is zero along the $x-$ and $y-$ axes, the partial derivatives $f_x$ and $f_y$ are both zero at the point $(0,0,0)$.

$f(x,y,z)$ is not even continuous at $(0,0,0)$ . Taking the limit along the line $x=y, z=0$ we have $\lim_{x\rightarrow 0}\frac{xx}{x^2+x^2}+c(0)=1/2+c(0)\ne f(0,0,0)$

Angela Pretorius
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