# Considering Existing of a Partial Derivative

1. Jul 3, 2010

### makyol

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I thought, in order to take derivative function must be cont. so it would be nice to check limf as (x,y) goes to (0, 0) but it did not seem to be a correct solution because appliying l'hopital gets harder and harder. If anyone can help me to solve this problem?

Thanks!

Last edited by a moderator: May 4, 2017
2. Jul 3, 2010

### HallsofIvy

Don't use L'hopital. Instead, convert to polar coordinates. That way, the single variable, r, measures the distance to the origin.

Since $x= r cos(\theta)$ and $y= r sin(\theta)$,

$$\frac{x^5+ y^6}{(x^2+ y^2)^\alpha}= \frac{r^5cos^5(\theta)+ r^6sin^6(\theta)}{r^\alpha}$$
$$= \frac{r^5}{r^\alpha}(cos^5(\theta)+ rsin^6(\theta)}$$

As long as $\alpha< 5$, that will go to 0 as r goes to 0 no matter what $\theta$ is and so the function will be continuous at (0, 0).

I would recommend you not worry about the continuity of the function itself but go ahead and take the derivatives with respect to x and y.