MHB Partial Derivatives: Solving at (0,0)

[Yankel]

Hello all,

I have this function here:

\[f(x,y)=\left\{\begin{matrix} z &(x,y)\neq (0,0) \\ 0 & (x,y)=(0,0) \end{matrix}\right.\]

where

\[z=\frac{x^{3}+xy^{2}}{2x^{2}+y^{2}}\]

And I need to find it's first partial derivative by x and y at the point (0,0). I am not sure I know how to approach this. At the point (0,0) the function is 0, so what's the "catch" ? The final answer is 0 by y and 0.5 by x, I got no clue how to get there...

Thanks !

 

[magneto1]

Re: Partial Derivaties

The catch is that $f$ has a point of discontinuity at $(0,0)$. To find the partial derivative at such point, you have to use the limit definition.

\[
\frac{\partial f}{\partial x} (0,0) = \lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h\to 0} \frac{1}{h} \cdot \frac{h^3}{2h^2} =
\lim_{h\to 0} \frac 12 = \frac 12
\]

You can work out $\frac{\partial f}{\partial y}$ by yourself.