Let $f:\mathbb{R^2} \to \mathbb{R}$ $$f(x) = \left\{ \begin{array}{ll} \frac{xy^2}{x^2+y^4}, & (x,y)\ne0 \\ 0, & (x,y) =0 \\ \end{array} \right.$$ Find the directional derivatives at the origin $D_af(0,0)$ for every direction $a=(a_1,a_2)$, when $||a||=1.$ Show that $f$ is not differentiable at the origin.

For the partials I found that $\frac{\partial}{\partial x} = \frac{y^2(y^4-x^2)}{(x^2+y^4)^2}$ and $\frac{\partial}{\partial y} = \frac{2xy(x^2-y^4)}{(x^2+y^4)^2}$

so $\nabla f=(\frac{y^2(y^4-x^2)}{(x^2+y^4)^2}, \frac{2xy(x^2-y^4)}{(x^2+y^4)^2})$.

The directional derivative is then $D_af=\nabla f\cdot a = (\frac{y^2(y^4-x^2)}{(x^2+y^4)^2}, \frac{2xy(x^2-y^4)}{(x^2+y^4)^2})\cdot(a_1,a_2)$

I'm not sure I understand what they mean by $D_af(0,0) = \nabla f(0,0)\cdot a$ this would lead to division by $0$ right?

Also for the differentiability I tried to use the definition of the partials and see if they're both continuous at the origin, but that lead to a very messy expression for instance $$\frac{\partial}{\partial x} = \frac{f(x+h,y)-f(x,y)}{h} = \frac{\frac{(x+h)y^2}{(x+h)+y^4}-\frac{xy}{x^2+y^4}}{h}$$

and this didn't seem to simplify to anything usable... What should I do here?

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    $\begingroup$ Hint: $f$ is not even continuous at the origin $\endgroup$ – Ninad Munshi Sep 23 '20 at 7:25
  • $\begingroup$ How can I show this? Simply looking at $x=0$ and $y=0$ separately didn't lead me to anything. Is there some kind of educated guess I should make to find directions which would lead to different limits? $\endgroup$ – user713999 Sep 23 '20 at 7:29
  • $\begingroup$ The formula $\mathrm D_af=\nabla f\cdot a$ only holds if $f$ is differentiable in the first place. You shouldn't use it if you're not sure wether $f$ is differentiable (and certainly not if the exercise tells you that it's not) $\endgroup$ – Vercassivelaunos Sep 23 '20 at 8:07
  • $\begingroup$ @Vercassivelaunos What other options do I have in order to find the directional derivative then? $\endgroup$ – user713999 Sep 23 '20 at 8:22
  • $\begingroup$ Use its definition: $\mathrm D_af(x)=\lim_{h\to 0}\frac{f(x +ah)-f(x)}{h}$. $\endgroup$ – Vercassivelaunos Sep 23 '20 at 8:41

We are requested to find the directional derivatives at the origin that is for $a\cdot b \neq 0$

$$\lim_{(ah,bh)\to(0,0)} \frac{\frac{ab^2h^3}{a^2h^2+b^4h^4}-0}{h} =\lim_{(ah,bh)\to(0,0)} \frac{ab^2h^3}{a^2h^3+b^4h^5}=\frac{b^2}a$$

with $f_x=f_y=0$.

For differentiability just note that for $x=y^2$

$$\lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2+y^4}=\lim_{(y^2,y)\to(0,0)} \frac{y^4}{y^4+y^4}=\frac12$$

therefore $f(x,y)$ is not continuous at the origin and then it is not differentiable.

  • $\begingroup$ Using the definition I came up with $\lim_{h\to 0}\frac{f(x +ha, y)-f(x, 0)}{h} = \lim_{h\to 0}\frac{f(0 + ha, 0)-f(0, 0)}{h} = \lim_{h\to 0}\frac{\frac{(ha)\cdot 0}{(ha)^2 + 0} - 0}{h} = \lim_{h\to 0} \frac{0}{h^2a^2} = 0$. How did you get a different result for this? $\endgroup$ – user713999 Sep 23 '20 at 9:49
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    $\begingroup$ @Daniel The definition for directinal derivatives at $(0,0)$ is $$\lim_{h\to 0}\frac{f(x +ha, y+hb)-f(0, 0)}{h} $$ $\endgroup$ – user Sep 23 '20 at 9:52

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