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Math Help - First and Second Partials at the Origin

  1. #1
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    First and Second Partials at the Origin

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

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

    Solving f_x(0,0) how I would normally take a partial derivative I get \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.
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  2. #2
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    Quote Originally Posted by Shananay View Post
    f(x,y)=\sqrt[5]{x^5+y^5}

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

    Solving f_x(0,0) how I would normally take a partial derivative I get \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 f_x(x,y) = \frac{x^4}{(x^5+y^5)^{\frac{4}{5}}} is valid whenever x+y\ne0. In particular, f_x(x,0) = \frac{x^4}{(x^5)^{\frac{4}{5}}}=1 when x\ne0. You have also shown (using the limit definition of the partial derivative) that  f_x(0,0) = 1.

    Therefore (again using the limit definition of the partial derivative) 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.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    \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?
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  4. #4
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    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 f_x at the points (x,0) and (0,0). I'm using the fact that the second derivative f_{xx} is the x-derivative of the first derivative.
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