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Math Help - Partial Derivative Problem

  1. #1
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    Partial Derivative Problem

    I have two functions
    r = g (x,y)
    s = h (x, y)
    so its given that

    x = r^3 - s
    y = s^3 -r

    g(0,0) = h(0,0) = 1

    partial derivitives (sorry wrong notation): dg/dx, dh/dy and d^2g/dx^2
    All at 0,0

    Thanks, any help would be appreciated
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by viraj View Post
    I have two functions
    r = g (x,y)
    s = h (x, y)
    so its given that

    x = r^3 - s
    y = s^3 -r

    g(0,0) = h(0,0) = 1

    partial derivitives (sorry wrong notation): dg/dx, dh/dy and d^2g/dx^2
    All at 0,0

    Thanks, any help would be appreciated
    Use the chain rule here..

    For \frac{\partial g}{\partial x}, we know by the chain rule that \frac{\partial g}{\partial x}=\frac{\partial g}{\partial r}\cdot\frac{\partial r}{\partial x}+\frac{\partial g}{\partial s}\cdot\frac{\partial s}{\partial x}

    We can figure out that \frac{\partial g}{\partial r}=1.

    To find \frac{\partial  r}{\partial x}, use partial implicit differentiation: x=r^3-s\implies 1=3r^2\frac{\partial r}{\partial x}\implies \frac{\partial r}{\partial x}=\frac{1}{3r^2}

    We also can figure out that \frac{\partial g}{\partial s}=0. We can disregard the \frac{\partial s}{\partial x} term, since it will end up disappearing.

    Thus, we now see that \frac{\partial g}{\partial x}=\left(1\right)\cdot\left(\frac{1}{3r^2}\right)=  \frac{1}{3r^2}

    Since, r=g(x,y), we see that \frac{\partial g}{\partial x}=\frac{1}{3(g(x,y))^2}

    Now evaluate it at (0,0). You should end up with g_x(0,0)=\frac{1}{3(g(0,0))^2}=\frac{1}{3(1)^2}=\c  olor{red}\boxed{\frac{1}{3}}

    ---------------------------------------------------------------------------------------

    Try to find \frac{\partial h}{\partial y} using the same technique as I used above.

    ---------------------------------------------------------------------------------------

    For finding \frac{\partial^2g}{\partial x^2}, recall that by the chain rule, we have \frac{\partial^2g}{\partial x^2}=\frac{\partial g}{\partial r}\cdot\frac{\partial^2r}{\partial x^2}+\frac{\partial^2 g}{\partial r^2}\cdot\left(\frac{\partial r}{\partial x}\right)^2 (I think I have this correct...if not, feel free to state what it should be)

    Follow the same process for this one as well.

    Does this make sense?

    --Chris
    Last edited by Chris L T521; November 16th 2008 at 11:12 PM.
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  3. #3
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    Chain Rule

    I don't think I understand the Chain Rules yet. Will go some reading and find out the exact theorem when it comes to partial derivatives
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