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I would like to find all directional derivatives of the function $$f(x,y) = (3x^4 + y^4)^{1/4} , $$ (where $ (x,y) \in \mathbb{R}^2 $), in the point $(0,0)$. I tried to do this by calculating $$\nabla f(x,y) = f_1 (x,y) e_1 + f_2 (x,y) e_2 $$, where $e_n$ is the $n$'th unit vector and $f_n$ is the partial derivative with respect to the $n$'th variable. I found that $f_1 (x,y) = (1/4)\cdot 12x^3 (3x^4 + y^4)^{-3/4} = \frac{3x^3}{(3x^4+y^4)^{3/4}} $, and that $f_2 (x,y) = \frac{y^3}{(3x^4 + y^4)^{3/4}} $. When filling in $(0,0)$ to find $\nabla f(0,0)$, however, I get two expressions involving a $\frac{0}{0}$ - fraction. I do think I need to do this though, since $$D_{v/|v|} f(a,b) = \langle (\frac{v}{|v|}, \nabla f(a,b) \rangle . $$ Can you please help me with this?

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up vote 2 down vote accepted

The intersection of the graph of $f$ with the plane $y=mx$ is the graph of the equation $y=(3+m^4)^{1/4}|x|$ (sketched in the plane $y=mx$). The directional derivative of $f$ at $(0,0)$ in the "direction of the plane" is the derivative of $y=(3+m^4)^{1/4}|x|$ evaluated at $0$ (or it's negative). But this derivative does not exist. So, the directional derivative of $f$ at $(0,0)$ does not exist in any direction.

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The origin is a very special point for this function. In particular, the partial derivatives there are of no help. Therefore we should go at these directional derivatives in the intuitive geometric sense; i.e. we should look at $${f(r\cos\phi, r\sin\phi)-f(0,0)\over r}=\root 4 \of {3 \cos^4\phi+\sin^4\phi}\ =: g(\phi)$$ and let $r\to 0+$. As the right side is in fact independent of $r$ the expression $g(\phi)$ is already the limit when $r\to 0+$: $$(D_\phi f)(0,0):=\lim_{r\to 0+}{f(r\cos\phi, r\sin\phi)-f(0,0)\over r}=\root 4 \of {3 \cos^4\phi+\sin^4\phi}\ ,$$ and it remains to plot it as a function of the polar angle $\phi$.

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When you look at $(x^4)^{1/4}$, you see that this is not differentiable at $x=0$. So I don't think that you can get a directional derivative at $(0,0)$ in your case.

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