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Considering Existing of a Partial Derivative

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://www.netbookolik.com/wp-content/uploads/2010/07/q1.png [Broken]

    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?

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


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    Science Advisor

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

    Since [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex],

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

    As long as [itex]\alpha< 5[/itex], that will go to 0 as r goes to 0 no matter what [itex]\theta[/itex] 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.
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