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Math Help - Showing that the directional derivative exists at a point.

  1. #1
    Senior Member Pinkk's Avatar
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    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 <a,b,c> 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.
    Last edited by Pinkk; September 20th 2010 at 08:28 PM.
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  2. #2
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    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.
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