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I was going to use the definition of derivative for $f : \Bbb{R}^n \to \Bbb{R}$ but when I was calculating $\nabla f(0, 0)$, it didn't exist, so now I'm stumped on what to do. Should I not plug in the point $(0, 0)$ into $\nabla f$ and then try to prove $$\lim_{h\to 0} \frac{f(a + h) - f(a) - \nabla f \cdot h}{||h||} \neq 0$$ for general $a$?

EtherealMist
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    Not just doesn't equal 0, but does not exist. – Doug M Jun 16 '17 at 01:08
  • There is a nice theorem stating that if $f$ is differentiable at a point $a$, then all the directional derivatives of $f$ exists at $a$. The converse is not true though. – edm Jun 16 '17 at 01:14

3 Answers3

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If $\nabla f (x_0)$ does not exist then $f$ is not differentiable at $x_0$.

Jason
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When one wants to prove some function is not differentiable, it is usually not very efficient to work on the definition. All one needs in your particular case is to look at some direction that $f$ has not derivative at $(0,0)$.

Hint:

  • Show that $f(x,x)=|x|$ is not differentiable at $x=0$.
  • Show that the statement above implies that $f$ is not differentiable at $(0,0)$.
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I might start by considering the partial derivatives at (x,y) = 0. We have $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\sqrt{\left|xy\right|}\right)$ or $\frac{\partial f}{\partial x} = c\frac{\partial}{\partial x}\left(\sqrt{\left|x\right|}\right)$ for $c=\sqrt{y}$, which is clearly not differentiable.

eric.mitchell
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