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I'm trying to compute the directional derivative at $(0,0)$ of $f$ in the direction $u=(u_1\ \ u_2)^T$ with $||u|| = 1$, where $f(x,y) =\begin{cases} \frac{x^3}{x²+y²} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0)\\ \end{cases}$

Now, I computed the partial derivatives of f at $(0,0)$, which I found to be respectively $1$ and $0$. Thus, I would expect the directional derivative in the direction $u=(u_1\ \ u_2)^T$ to be $(1 \ \ 0) \cdot (u_1\ \ u_2)^T = u_1$. However, according to the answer sheet, the answer should be $u_1^3$. What am I doing wrong?

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1 Answer 1

up vote 2 down vote accepted

What you're doing wrong is assuming that $f$ is smoothly differentiable at $(0,0)$. Instead of using the gradient shortcut to calculate the directional derivative, you should just use the definition of directional derivatives. (By the way, as a practical note, you haven't used the hypothesis that $\|u\| = 1$.)

I'll remind you of that definition. For convenience's sake, write $f$ as a function of vectors $(x,y) = \mathbf{x}$. The directional derivative of $f$ at $\mathbf{x}$ in the direction of $\mathbf{v}$ is $$\lim_{t\to 0}\frac{1}{t} \bigg( f(\mathbf{x} + t\mathbf{v}) - f(\mathbf{x})\bigg).$$ So now do the only thing you can: plug in for $\mathbf{x}$, $\mathbf{v}$, and compute the limit.

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Thanks! Somehow that definition isn't in my notes or book, so I didn't think of using it (though I probably should have thought of looking up the definition on the internet). –  Max Feb 15 '13 at 4:21

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