Results 1 to 10 of 10

Math Help - Evaluating Partial Derivative

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    12

    Evaluating Partial Derivative

    The question is:
    Evaluate f_{xy}(0, 0) and f_{yx}(0, 0) for f(x, y) = xy\frac{x^2−y^2}{x^2+y^2}.
    What would you say about Clairaut's theorem?

    I found fxy(x,y) but then I realized... wouldn't this be undefined if you evaluate at f(0,0)?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    If you read the text of Clairaut's theorem carefully, you might see what the problem is getting at. What did you get for evaluating the mixed partial derivatives at the origin?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    12
    Quote Originally Posted by Ackbeet View Post
    If you read the text of Clairaut's theorem carefully, you might see what the problem is getting at. What did you get for evaluating the mixed partial derivatives at the origin?
    For f_{xy}=\frac{x^8-y^8}{(x^2+y^2)^4} Is it not undefined if evaluated at the origin? Does this mean that the theorem doesn't apply?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Hmm. Not sure you took the derivative correctly. Just to be clear, the function is defined by

    f(x,y)=xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}.

    Is that correct? If so, what do you get, precisely, for f_{x}(x,y)? And then what do you get for f_{xy}(x,y)? Please show your steps.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2008
    Posts
    12
    f(x,y) = \frac{x^3y - xy^3}{x^2+y^2}

    f_x(x,y) = \frac{(3x^2y-y^3)(x^2+y^2) - 2x(x^3y-xy^3)}{(x^2+y^2)^2}<br />
                = \frac{x^4y+4x^2y^3-y^5}{(x^2+y^2)^2}

    f_{xy}(x,y) = \frac{(x^4+12y^2x^2-5y^4)(x^2+y^2)^2 - 2(x^2+y^2)(2y(x^4y+4x^2y^3-y^5)}{(x^2+y^2)^4}
    = \frac{(x^8+14x^6y^2+20x^4y^4+2x^2y^6-5y^8) - (4x^6y^2+20x^4y^4+14x^2y^6-4y^8)}{(x^2+y^2)^4}
    Oh, I guess I messed up here (assuming I did everything above correctly, anyways)

    = \frac{x^8+10x^6y^2-12x^2y^6-y^8}{(x^2+y^2)^4}

    But still, wouldn't the denominator become 0 at the origin?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Yes, the denominator will become 0 at the origin. This means Clairaut's theorem does not apply. However, I think you'll find that the mixed partial derivatives have the same form. So what is that telling you?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2008
    Posts
    12
    Hmm... Does this mean the results of the theorem can still occur when the theorem itself doesn't apply?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Yes. Think about the statement, "If A then B." This is the format of most theorems. B can be true whether or not A is or not. The situation you cannot have is when A is true and B is false. Any other truth combination of A and B can still hold.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    May 2008
    Posts
    12
    Quote Originally Posted by Ackbeet View Post
    Yes. Think about the statement, "If A then B." This is the format of most theorems. B can be true whether or not A is or not. The situation you cannot have is when A is true and B is false. Any other truth combination of A and B can still hold.
    Ah, it's nice and clear now. Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You're very welcome. Have a good one!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative of arctan in a partial derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 12th 2010, 01:52 PM
  2. Evaluating a First Partial at a Point
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 8th 2010, 04:56 AM
  3. Evaluating a Derivative at a Given point
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 14th 2009, 01:27 PM
  4. partial derivative please help
    Posted in the Calculus Forum
    Replies: 15
    Last Post: April 23rd 2008, 12:31 PM
  5. Evaluating Partial Fractions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 9th 2007, 04:43 PM

Search Tags


/mathhelpforum @mathhelpforum