# Showing that the directional derivative exists at a point.

• Sep 20th 2010, 07:37 PM
Pinkk
Showing that the directional derivative exists at a point.
Show that for any unit vector $\displaystyle u$, the directional derivative at $\displaystyle 0$ in the direction of $\displaystyle u$ exists for the function $\displaystyle f(x,y,z) = (xyz)^{1/3}$.

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

Edit: So I realized I had to use the limit definition of directional derivative, and I got for a unit vector $\displaystyle <a,b,c>$ the directional derivative at $\displaystyle 0$ is $\displaystyle (abc)^{1/3}$ and since the gradient is undefined at $\displaystyle 0$, the gradient dotted with the unit vector is not $\displaystyle (abc)^{1/3}$ so the function is not differentiable at $\displaystyle 0$. Is this right? And for a second part where $\displaystyle f(x,y,z)=(x^{2}yz)^{1/3}$ instead, I get that the directional derivative is $\displaystyle 0$ and that once again, the gradient is undefined at $\displaystyle 0$, so the gradient dotted with the unit vector is not $\displaystyle 0$, so the function is not differentiable at $\displaystyle 0$.
• Sep 20th 2010, 09:47 PM
Jose27
Remember the definition of directional derivative in the direction $\displaystyle u: \partial_uf(x)= \lim_{h\rightarrow 0} \frac{f(x+hu)-f(x)}{h}$ where in your case is $\displaystyle x=(0,0,0)$ and $\displaystyle u=(u_1,u_2,u_3)$ 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.