# Why is this function not differentiable at $(0,0)$

I was brushing up on some calculus and I was thinking about the following function:

Let$$f(x) = \begin{cases} \frac{yx^{6} +y^{3}+x^{3}y}{x^{6}+y^{2}} & \text{for (x,y) \neq (0,0),} \\ 0 & \text{for (x,y) = (0,0). } \\ \end{cases}$$The function $f$ is continuous over the entire real line and is differentiable everywhere except at $x=0$.

For some arbitrary nonzero $\xi=(u,v)$, I was able to compute the directional derivative ${\xi}f(0,0)=\frac{1}{h}f(hu,hv)=v$. In particular for $\xi=\frac{\partial}{\partial{x}}=(1,0)$, we have $\frac{\partial{f}}{\partial{x}}(0,0)=0$ and for $\xi=\frac{\partial}{\partial{y}}=(0,1)$, we have that $\frac{\partial{f}}{\partial{y}}(0,0)=1$. From here we see that partial derivative depends linearly on the vector we differentiate on meaning $(t{\xi}))f(0,0)=(tu,tv)f(0,0)=tv=t({\xi}f(0,0))$.

This function is not differentiable at the origin, in fact it is not even continuous at the origin. I was having trouble seeing exactly why. I've forgotten lots of calculus, but I was wondering if the fact that $\frac{\partial{f}}{\partial{y}}(0,0)=1$ while $\frac{\partial{f}}{\partial{x}}(0,0)=0$ implies discontinuity at $(0,0)$.

Also as a side note, there is the big theorem in differential calculus which says if all the directional derivatives ${\xi}f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ are continuous at a point $p$, then $f$ is differentiable at the point $p$. I'm confused as to how the partial derivatives above are not continuous.

Any help is much appreciated.

• Aug 14, 2018 at 23:49

Hint: consider approaching the origin along the path $(x,x^3)$

• The path $(x,x)$ will also be revealing.
– zhw.
Aug 8, 2015 at 2:24
• Thanks a lot for the hint. I posted an answer below just for practice that incorporates the hint. Thanks again. Aug 10, 2015 at 0:59
• @user135520 You're welcome :) Aug 10, 2015 at 1:35

We have that differentiability implies continuity, therefore if $f$ is differentiable at the origin then it will be continuous at the origin. Which means we must have

$\lim_{(x,y) \to (0,0)} f(x,y)=(0,0)$

We consider approaching the origin along the path $(x,x^{3})$ where

$\lim_{(x,y) \to (0,0)} \frac{yx^{6} +y^{3}+x^{3}y}{x^{6}+y^{2}}$ becomes

$\lim_{x \to 0} \frac{2x^{9}+x^{6}}{x^{6}}=\frac{1}{2}$

This implies that $\lim_{(x,y) \to (0,0)} f(x,y) \neq (0,0)$ which implies that $f$ is not continuous.