Results 1 to 4 of 4

Math Help - is f differentiable

  1. #1
    Banned
    Joined
    Nov 2008
    Posts
    63

    is f differentiable

     f(x,y)=\frac{x^{2}y+xy^{2}}{x^{2}+y^{2}} , if \ \ (x,y)\neq(0,0)<br />
and \ \ 0 \ \ otherwise<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,984
    Thanks
    1647
    Quote Originally Posted by silversand View Post
     f(x,y)=\frac{x^{2}y+xy^{2}}{x^{2}+y^{2}} , if \ \ (x,y)\neq(0,0)<br />
and \ \ 0 \ \ otherwise<br />
    It is clearly differentiable, by the usual differentiation formulas, for any point other than (0,0):
    \frac{\partial f}{\partial x}= \frac{(2xy+ y^2)(x^2+ y^2)- (x^2y+ xy^2)(2x)}{x^2+ y^2}= \frac{2xy^3+ y^4- x^2y^2}{(x^2+ y^2)^2}
    \frac{\partial f}{\partial y}= \frac{(x^2+ 2xy)(x^2+ y^2)- (x^2y+ xy^2)(2y)}{x^2+ y^2}= \frac{x^4+ 2xy^3- x^2y^2}{(x^2+ y^2)^2}
    Since the partial derivatives are continuous, the function is differentiable. Further, the limits, as (x, y) go to 0, are 0.

    So the only question is at (0,0).
    \frac{f(h,0)- f(0,0)}{h}= 0/h= 0
    \frac{f(0,h)- f(0,0)}{h}= 0/y= 0

    Since those match the limits of the partial derivatives as (x, y) goes to (0,0), the function is also differentiable at (0,0).

    Again, that shows that the partial derivatives exist and are continuous at (0,0). That implies that the function is differentiable at (0,0).
    Last edited by HallsofIvy; May 10th 2009 at 07:10 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Nov 2008
    Posts
    63
    I am not sure with the continuous bit.

    as (x,y) goes to 0, the denominator should be zero?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,984
    Thanks
    1647
    Quote Originally Posted by silversand View Post
    I am not sure with the continuous bit.

    as (x,y) goes to 0, the denominator should be zero?
    As far as (x,y) not (0,0) is concerned, the partial derivatives give rational function with NON-zero denominator. A rational function is always continuous at such a point.

    As far as (0,0) is concerned, I showed that the partial derivatives at (0,0) are both 0.

    To see what the limit of the derivatives is at (0,0), it is best to change to polar coordinates.

    \frac{\partial f}{\partial x}=\frac{2xy^3+ y^4- x^2y^2}{(x^2+ y^2)^2}
    In polar coordinates, x= r cos(\theta), y= r sin(\theta) and r^2= x^2+ y^2
    The derivative becomes
    \frac{2r^4 cos(\theta)sin^3(\theta)+ r^4sin^4(\theta)- r^4cos^2(\theta)sin^2(\theta)}{r^2} = r^2(cos(\theta)sin^2(\theta)+ sin^4(\theta)- cos^2(\theta)sin^2(\theta))
    Because of the r^2 factor, the limit, as r goes to 0, is 0, no matter what \theta is. Because, in polar coordinates, r alone measures the distance from (0,0), that is enough to show that the limit is 0.

    I'll let you do the same for the other partial derivative.
    Last edited by HallsofIvy; May 10th 2009 at 07:20 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: May 15th 2011, 02:55 AM
  2. Replies: 9
    Last Post: December 17th 2010, 08:13 AM
  3. Replies: 0
    Last Post: October 3rd 2010, 07:03 AM
  4. Replies: 1
    Last Post: February 7th 2010, 12:18 AM
  5. Sequence of differentiable functions, non-differentiable limit
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 3rd 2009, 05:13 AM

Search Tags


/mathhelpforum @mathhelpforum