# Directional Derivative Question

• dr721

## Homework Statement

There exists no function f: ℝn $\rightarrow$ ℝ so that for some point a $\in$ ℝn, we have Dvf(a) > 0 for all nonzero vectors v $\in$ ℝn.

2. The attempt at a solution

No quite sure where to go with this one, any help would be great!

Consider the derivative along some v and along -v.

Right, so that'd be D and -D as -v is multiplied by the scalar -1. Would it be possible for somebody to restate the question? I'm not sure I understand what it's asking.

Well, its your question! How can someone else tell you what you are asking. You have probably copied it incorrectly because, as given, it makes no sense. You say "Dv(a)> 0" but you cannot "order" a vector space so it also makes no sense to say that a vector is "larger" than the 0 vector.

Yes, it is my question, and it is also copied verbatim from our book. And this question is not listed under the list of errors for the book, so I'm sorry if it seems wrong, but I have a strong feeling knowing my professor (who authored the book) that it is indeed accurate.

Nevermind, I just thought about that v, -v statement. Thanks voko!

## Homework Statement

There exists no function f: ℝn $\rightarrow$ ℝ so that for some point a $\in$ ℝn, we have Dvf(a) > 0 for all nonzero vectors v $\in$ ℝn.

2. The attempt at a solution

No quite sure where to go with this one, any help would be great!

The statement is false as written, assuming that Dvf(a) means the directional derivative of f at a in direction v. A simple 1-dimensional counterexample is f(x) = |x|, whose directional derivative at x = 0 is +1 in both directions +1 and -1. (Of course, |x| is not differentiable at x = 0, but it _does_ have directional derivatives there.) A 2-dimensional counterexample is f(x,y) = √(x2+y2). Again, this has positive directional derivatives at (x,y) = (0,0), but is not a differentiable function there. Are you sure you have stated the problem correctly?

RGV

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