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Let $f(x,y)= (xy)^{1/3}$

Using definition of the partial derivatives, prove $f_x(0,0), f_y(0,0)$ both exist.

Show that the directional derivative for $f$ in any direction other than $i$ or $j$ does not exist at the origin.

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  • $\begingroup$ Have you tried writing down what the definition of the partial derivative looks like in this case? $\endgroup$ – Paul Siegel Mar 12 '14 at 19:00
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I assume that for $t\in{\mathbb R}$ the expression $t^{1/3}$ is defined as $\ {\rm sgn}(t)\>|t|^{1/3}$.

It follows that $${f(x,0)-f(0,0)\over x}=0\qquad(x\ne0)\ ,$$ which implies $f_x(0,0)=0$; and similarly $f_y(0,0)=0$.

Let $u:=(\cos\phi,\sin\phi)$ be a fixed unit vector. Then by definition the directional derivative of $f$ at $(0,0)$ in direction $u$ is given by $$\eqalign{D_uf(0,0)&=\lim_{t\to 0+}{f(t\cos\phi,t\sin\phi)-f(0,0)\over t}= \lim_{t\to 0+}{t^{2/3}|\cos\phi\>\sin\phi|^{1/3}\over t}\cr &=\lim_{t\to 0+}{|\cos\phi\>\sin\phi|^{1/3}\over t^{1/3}}\ ,\cr}$$ if the limit on the right hand side exists. But when $\phi$ is not an integer multiple of ${\pi/2}$ this limit does not exist.

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