# Prove partial derivatives exist, but not all directional derivatives exists.

During my analysis course my teacher explained the difference between partial derivatives and directional derivatives using the notion that a partial derivatives looks at the function as approaching a point along the axes (in case of of the plane), and a directional derivative as approaching a point from any direction in the plane. He also explained that the existence of directional derivatives is a stronger notion than the existence of partial derivatives exists: if all directional derivatives exist, then the partial derivatives exist too.

I am to show (not necessarily prove) a case where the partial derivatives exist, but not all directional derivatives exist (hence, f is not differentiable).

• note that even if all directional derivatives exists, $f$ need not to be differentiable :-) – Ant Apr 19 '16 at 15:32
• Thank you. I'm meant to show this in a later exercise too. For now I only want to show an example I think. I can't come up with a function, however. – falidoro Apr 19 '16 at 15:35

Consider for instance $$f(x,y)=\begin{cases} \frac{xy}{x^2+y^2}&\text{if }(x,y)\neq(0,0),\\ 0&\text{otherwise.}\end{cases}$$ The partial derivatives exist everywhere. Away from the origin this is clear. At the origin we need to calculate the partials from first principles: $$\frac{f(0+h,0)-f(0,0)}h=\frac{0-0}h=0$$ so letting $h\to0$ we find $f_x(0,0)=0$. By symmetry we also have $f_y(0,0)=0$. However, not all directional derivatives exist at the origin. For example, let $\mathbf v=(\frac1{\sqrt{2}},\frac1{\sqrt{2}})$. Then $$\frac{f(\mathbf0+t\mathbf v)-f(\mathbf0)}{t}=\frac{\frac{t^2}{t^2+t^2}-0}t=\frac1{2t}$$ which does not have a limit as $t\to0$.
• Usually functions of the form $f(x,y)=\frac{x^jy^k}{(x^2+y^2)^n}$ for $(x,y)\neq(0,0)$ are good for these kinds of questions. Tweak the exponents depending on whether you're looking at continuity or differentiability. – Jason Apr 19 '16 at 18:40