Results 1 to 4 of 4

Thread: First and Second Partials at the Origin

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    62

    First and Second Partials at the Origin

    $\displaystyle f(x,y)=\sqrt[5]{x^5+y^5}$

    The question is to find $\displaystyle f_x(0,0)$ and $\displaystyle f_{xx}(0,0)$

    Solving $\displaystyle f_x(0,0)$ how I would normally take a partial derivative I get $\displaystyle \frac{x^4}{(x^5+y^5)^{\frac{4}{5}}}$ which is undefined at (0,0).

    I get 1 when I use the limit definition of the partial derivative. I'm unsure how to calculate the the second partial though. I want to get an actual number and not 0/0 if possible.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Shananay View Post
    $\displaystyle f(x,y)=\sqrt[5]{x^5+y^5}$

    The question is to find $\displaystyle f_x(0,0)$ and $\displaystyle f_{xx}(0,0)$

    Solving $\displaystyle f_x(0,0)$ how I would normally take a partial derivative I get $\displaystyle \frac{x^4}{(x^5+y^5)^{\frac{4}{5}}}$ which is undefined at (0,0).

    I get 1 when I use the limit definition of the partial derivative. I'm unsure how to calculate the the second partial though. I want to get an actual number and not 0/0 if possible.
    The formula $\displaystyle f_x(x,y) = \frac{x^4}{(x^5+y^5)^{\frac{4}{5}}}$ is valid whenever $\displaystyle x+y\ne0$. In particular, $\displaystyle f_x(x,0) = \frac{x^4}{(x^5)^{\frac{4}{5}}}=1$ when $\displaystyle x\ne0$. You have also shown (using the limit definition of the partial derivative) that$\displaystyle f_x(0,0) = 1$.

    Therefore (again using the limit definition of the partial derivative) $\displaystyle f_{xx}(0,0) = \lim_{x\to0}\frac{f_x(x,0)-f_x(0,0)}x = \lim_{x\to0}\frac{1-1}x = \lim_{x\to0}\,\frac0x = 0$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    62
    Quote Originally Posted by Opalg View Post
    $\displaystyle \lim_{x\to0}\frac{f_x(x,0)-f_x(0,0)}x = \lim_{x\to0}\frac{1-1}x$
    I'm not sure I understand this step. What exactly are the functions in the first numerator?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Shananay View Post
    I'm not sure I understand this step. What exactly are the functions in the first numerator?
    They are the values of the first x-derivative $\displaystyle f_x$ at the points (x,0) and (0,0). I'm using the fact that the second derivative $\displaystyle f_{xx}$ is the x-derivative of the first derivative.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inequality of mixed partials
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Jan 5th 2010, 10:25 PM
  2. Help me please.... Mixed Partials
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Sep 18th 2009, 08:02 AM
  3. 2nd partials test
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Apr 2nd 2009, 01:56 PM
  4. Second partials
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 28th 2009, 12:20 PM
  5. [SOLVED] First-Order partials
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Sep 29th 2008, 07:48 PM

Search Tags


/mathhelpforum @mathhelpforum