# Proving that a function is not differentiable using a certain definition

I have been given the following function:

$$f(x,y) = \begin{cases} \dfrac{xy(2x^2 - y^2)}{2x^2 + y^2} & \text{if (x, y) \ne (0, 0)} \\ 0 & \text{if (x, y) = (0, 0)} \end{cases}$$

Now we would like to show that the partial derivatives are both continuous at the point $(0, 0)$, but neither one of them is differentiable at (0, 0). First we should analyze the behavior at $(0, 0)$, as the quotient rule will not apply at that point, so we should go by the definition:

$$f_x(0, 0) = \lim_{x \rightarrow 0} \dfrac{f(x, 0) - f(0, 0)}{x} = \lim_{x \rightarrow 0} \dfrac{0 - 0}{x} = 0$$ $$f_y(0, 0) = \lim_{y \rightarrow 0} \dfrac{f(0, y) - f(0, 0)}{y} = \lim_{y \rightarrow 0} \dfrac{0 - 0}{y} = 0$$

Then we can obtain the following partial derivatives:

$$\frac{\partial f}{\partial x} = \begin{cases} \dfrac{4 x^4 y + 8 x^2 y^3 - y^5}{\left(2x^2 + y^2\right)^2} & \text{if (x, y) \ne (0, 0)} \\ 0 & \text{if (x, y) = (0, 0)} \end{cases}$$

$$\frac{\partial f}{\partial y} = \begin{cases} \dfrac{4x^5 - 8x^3 y^2 - xy^4}{\left(2x^2 + y^2 \right)^2} & \text{if (x, y) \ne (0, 0)} \\ 0 & \text{if (x, y) = (0, 0)} \end{cases}$$

My goal is to show that both of them are continuous at $(0, 0)$, but neither is differentiable. I have already shown that both of them are continuous by showing that both partials have a limit of $0$ at $(0, 0)$. (I just took the limit in polar coordinates).

But to show differentiability, I am to use the following definition of differentiability:

A function $f(x, y)$ is differentiable at $(x_0, y_0)$ if it can be expressed in the form:

$$f\left(x, y\right) = f\left(x_0, y_0\right) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \xi \left(x, y\right)$$

Where $\Delta x = x - x_0$, $\Delta y = y - y_0$, $r = \sqrt{(x - x_0)^2 + (y - y_0)^2}$ and $\xi$ has the following property:

For any $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $\delta < r$, $\left | \xi \right | < \epsilon r$.

In other words, $\left | \xi \right |$ disappears faster than any linear multiple of $r$ as the point is approached. I am to use this definition to prove that the partial derivatives are not differentiable. How could I prove that, for some $\epsilon > 0$, there is no valid choice of $\delta$?

• "Now we would like to show that the partial derivatives are both continuous at the point (0,0) but neither one of them is continuous." What?
– zhw.
Commented Nov 5, 2015 at 19:11
• Fixed (I meant differentiable) @zhw Commented Nov 5, 2015 at 19:13
• "Now we would like to show that the partial derivatives are both continuous at the point (0,0) but neither one of them is differentiable." You should say "neither one of them is differentiable at (0,0)".
– zhw.
Commented Nov 5, 2015 at 19:15
• Incidentally, even though I'm pointing out a few errors, this is an excellent write-up of a question
– zhw.
Commented Nov 5, 2015 at 19:22

Hint: You can see that $f_x(x,0)\equiv 0,$ so $f_{xx}(0,0) = 0.$ Also $f_x(0,y) = -y.$ Thus $f_{xy}(0,0) = -1.$ Now if $f_x$ were differentiable at $(0,0),$ we would have

$$f_x(x,y) = f_x(0,0) + f_{xx}(0,0)\cdot x + f_{xy}(0,0)\cdot y + \xi = -y + \xi.$$

Is this true? (Consider $f_x(x,x)$ for example.)

• But doesn't this mean that $\xi$ is $\frac{20x}{9}$ along the line y = x? Why is this disallowed? Commented Nov 5, 2015 at 20:00
• Well if had differentiability, then $\xi(x,y)/(x^2+y^2)^{1/2} \to 0$ but $\xi(x,x)/(x^2+x^2)^{1/2} \not \to 0.$
– zhw.
Commented Nov 5, 2015 at 20:50
• I see. So how would I formally present this argument using the above "epsilon delta" definition? Commented Nov 5, 2015 at 20:52
• The $\epsilon - \delta$ business says precisely $\xi/r \to 0,$ but we've just shown that isn't true.
– zhw.
Commented Nov 5, 2015 at 20:56