13
$\begingroup$

I have this function $$ f(x,y) = \left\{ \begin{array}{ll} \frac{x^3}{x^2 + y^2} & \mbox{if } (x,y) \neq (0,0) \\ 0 & \mbox{if } (x,y) = (0,0) \end{array} \right. $$ And I want to find the directional derivative in the $(1,1)$ direction at $(0,0)$, so using the limit definition this gives me $$ \begin{align*} D_vf(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(1,1)) - f(0,0)}{t} = \lim_{t\to 0}\frac{t^3}{2t^2 t} = \frac{1}{2} \end{align*} $$ But wolfram alpha is giving me $0$ as a result Wolfram alpha result
Is there something wrong with my procedure?
Any help appreciated :)

$\endgroup$
14
$\begingroup$

You are right and Wolfram is wrong. What happens is that the computer is using the gradient to calculate the directional derivative. But the formula to calculate the directional derivative using the gradient uses the chain rule, which assumes the function to be differentiable. Your function is not differentiable at $(0,0)$, so the gradient cannot be used to find the directional derivatives there.

It is worth mentioning that WA is making a second mistake: since the partial derivatives are not continuous at $(0,0)$, they cannot be calculated using differentiation rules (and I don't really know how they get the value zero). Calculating explicitly, $$ \frac{\partial f}{\partial x}f(0,0)=\lim_{h\to 0}\frac{h^3}{h(h^2+0)}=1, $$ $$ \frac{\partial f}{\partial y}f(0,0)=\lim_{h\to 0}\frac{0}{h(0+h^2)}=0. $$ Still, the formula $(a,b)\cdot\nabla f(0,0)=a$ gives the wrong result because $f$ is not differentiable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.