# Thread: Showing that the directional derivative exists at a point.

1. ## Showing that the directional derivative exists at a point.

Show that for any unit vector $u$, the directional derivative at $0$ in the direction of $u$ exists for the function $f(x,y,z) = (xyz)^{1/3}$.

Okay, so maybe I'm just rusty on my multivariable calculus, but aren't each of the partial derivatives undefined at $(0,0,0)$? What am I missing here? And I'm missing how I would prove whether or not the function is actually differentiable at $(0,0,0)$.

Edit: So I realized I had to use the limit definition of directional derivative, and I got for a unit vector $$ the directional derivative at $0$ is $(abc)^{1/3}$ and since the gradient is undefined at $0$, the gradient dotted with the unit vector is not $(abc)^{1/3}$ so the function is not differentiable at $0$. Is this right? And for a second part where $f(x,y,z)=(x^{2}yz)^{1/3}$ instead, I get that the directional derivative is $0$ and that once again, the gradient is undefined at $0$, so the gradient dotted with the unit vector is not $0$, so the function is not differentiable at $0$.

2. Remember the definition of directional derivative in the direction $u: \partial_uf(x)= \lim_{h\rightarrow 0} \frac{f(x+hu)-f(x)}{h}$ where in your case is $x=(0,0,0)$ and $u=(u_1,u_2,u_3)$ now just substitute and evaluate.

Another thing is that this particular function gives a counterexample to the converse of the theorem in your other thread.