# Showing that the function given by $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ and $f(0,0)=0$ is continuous but not differentiable

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?

• Are you sure the partial derivatives exist? Dec 28, 2013 at 18:10
• Yes, showing the function is differentiable is the same as showing that its derivative exists. How are you defining the derivative of a function $f:\mathbb R^2\to\mathbb R$? Dec 28, 2013 at 18:11
• @AndresCaicedo The partial derivatives (even all directional derivatives) may exist without the function being differentiable. Dec 28, 2013 at 18:13
• @egreg Yes.${}$ Dec 28, 2013 at 18:16
• 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) Dec 28, 2013 at 20:12

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.

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 $z=f(x,y)$.

Since $$|f|\le|x|$$, $$\lim_{(x,\,y)\to(0,\,0)}f=0$$. But $$\frac{\partial f}{\partial x}=\frac{y^3}{(x^2+y^2)^{3/2}}$$, which $$\to1$$ along an $$x=0,\,y\to0$$ path, or $$\to0$$ along an $$x\to0,\,y=0$$ path.

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)

• You can add it in your question :) Dec 29, 2013 at 4:56
• Why is this not of that form? That seems to be precisely what you need to show. Dec 29, 2013 at 6:44
• 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 Dec 29, 2013 at 9:20

CHECK IT, it's easier one! U must determine the partial derivatives first.

function of two Variable $$f(x, y)$$ is disterentiable at point $$(a,b)$$ it $$(h, k) \rightarrow(0,0) \quad \frac{f(a+h, b+k)-f(a, b)-h f_{x}(a, b)-k f_{y}(a, b)}{\sqrt{h^{2}+k^{2}}}$$ $$f_{x} \rightarrow$$ partial derivative wrt $$x$$ i.e $$f(x, y)=\frac{x y}{\sqrt{x^{2}+y^{2}}} ; \quad \quad f_{x}=\operatorname{lt} \frac{f(x, 0)-f(0,0)}{x}$$ \begin{aligned}(x, b) \rightarrow(0,0) & \text { at }(0,0) \\ \text { accordingly } & f_{y}=0 \end{aligned} lt$$(h, k) \rightarrow(0,0) \quad \frac{f(h, k)-f(0,0)-h f_{x}(0,0)-k f_{y}(0,0)}{\sqrt{h^{2}+k^{2}}}$$ $$=\frac{\frac{h K}{\sqrt{h^{2}+K^{2}}} \ -0-0-0}{\sqrt{h^{2}+K^{2}}}=\frac{h K}{\left(h^{2}+K^{2}\right)}$$ along the path $$k=m h$$ $$\lim _{h \rightarrow 0} \frac{m h \cdot h}{h^{2}+m^{2} h^{2}}=\log _{k \rightarrow 0} \frac{m h^{2}}{h^{2}\left(1+m^{2}\right)}=\frac{m}{1+m^{2}}$$ Limiting value is not unique. Hence not differentiable though it is continous