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Let $$ f(x,y) = \begin{cases} \dfrac{xy}{\sqrt{x^2+y^2}} & \text{if $(x,y)\neq(0,0)$ } \\[2ex] 0 & \text{if $(x,y)=(0,0)$ } \\ \end{cases} $$ Show that this function is continuous but not differentiable at $(0,0),$ although it has both partial derivatives existing there.

I can show this function is continous and the partial derivatives exist. But how can I show that this function is not differentiable?

Is showing that the function is differentiable similar to showing that a derivative exists?

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  • $\begingroup$ Are you sure the partial derivatives exist? $\endgroup$ – egreg Dec 28 '13 at 18:10
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    $\begingroup$ Yes, showing the function is differentiable is the same as showing that its derivative exists. How are you defining the derivative of a function $f:\mathbb R^2\to\mathbb R$? $\endgroup$ – Andrés E. Caicedo Dec 28 '13 at 18:11
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    $\begingroup$ @AndresCaicedo The partial derivatives (even all directional derivatives) may exist without the function being differentiable. $\endgroup$ – egreg Dec 28 '13 at 18:13
  • $\begingroup$ @egreg Yes.${}$ $\endgroup$ – Andrés E. Caicedo Dec 28 '13 at 18:16
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    $\begingroup$ I showed the partial derivatives exist by using polar coordinates.Still I do not understand how to show that this is not differentiable. Can I use the theorem " If f is continuous at (a,b) and the partial derivatives are defined and continuous at (a,b) , then f is differentiable at (a,b)". But isn't this a sufficient condition and it is not necessary right? (That is can't there be a situation where we are unable to show that the partial derivatives are not continuous but yet we find that the function is differentiable) $\endgroup$ – clarkson Dec 28 '13 at 20:12
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There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.

Graph

When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.

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I don't know if I should write my answer here but can't write all this as a comment.
This is what I tried to do:
According to the definition , $f$ is fifferentiable at $a $ if,
$\Delta$f=$f$(a$_1$+h$_1$,...,a$_n$+h$_n$)-$f$(a$_1$,a$_2$,..,a$_n$)
=l$_1$h$_1$+l$_2$h$_2$+...+l$_n$h$_n$+...+B$_1$h$_1$+...+B$_n$h$_n$
where l's are constants and $ \lim_{h1h2...hn\to 0,0,,..,0}$B$_i$=0

Here a=(0,0,).
Therefore $\Delta$f=f(0+h$_1$,0+h$_2$)-f(0,0)
=$\frac{h_1h_2}{\sqrt{h_1^2+h_2^2}}$-0

This is not in the above form .Therefore f is not differentiable at (0,0)

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  • $\begingroup$ You can add it in your question :) $\endgroup$ – Shuchang Dec 29 '13 at 4:56
  • $\begingroup$ Why is this not of that form? That seems to be precisely what you need to show. $\endgroup$ – Andrés E. Caicedo Dec 29 '13 at 6:44
  • $\begingroup$ Because Delta f can't be written as l$_1$h$_1$ +l$_2$h$_2$.If this can be written in that form then I am showing that the function is differentiable right?What I have to show is that the function is not differentiable $\endgroup$ – clarkson Dec 29 '13 at 9:20

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