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Math Help - Second derivatives using implicit differentiation

  1. #1
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    Second derivatives using implicit differentiation

    Hi there. Well, I wanted to know how to find the second derivatives of a function using implicit differentiation. Is it possible? I think it is. I think I must use the chain rule somehow, but I don't know how... I'm in multivariable calculus since the function I'm gonna use could be seen as a function of only one variable.

    An ellipse: F(x,y)=4x^2+y^2-25=0

    So we have the partial derivatives:
    F_x=8x, F_y=2y
    F_{xx}=8, F_{yy}=2

    So then, using implicit differentiation:

    \frac{{\partial x}}{{\partial y}}=-\displaystyle\frac{\frac{{\partial F}}{{\partial y}}}{\frac{{\partial F}}{{\partial x}}}=\displaystyle\frac{-y}{4x}

    But now if I wanna find \frac{{\partial^2 x}}{{\partial y^2}} how should I proceed?
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  2. #2
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    Well, if you actually want to find \frac{\partial^2 x}{\partial y^2} and not \frac{\partial^2 y}{\partial x^2}, then you will just differentiate the derivative you already have, i.e.
    \frac{\partial^2 x}{\partial y^2} = \frac{\partial x}{\partial y}\left[-\frac{y}{4x}\right]
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  3. #3
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    The thing is that it gives -\frac{1}{4x}, and If we differentiate expressing x=x(y)

    Look, this is the second derivative using the explicit form: Clic here
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  4. #4
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    I do this differently than you, so hopefully I won't confuse you.
    Here I am assuming that x is a function of y.
    \frac{d}{dy}\left[4x^2+y^2-25\right]=\frac{d}{dy}\left[0\right]
    8xx'+2y=0 \Rightarrow x' = -\frac{y}{4x}
    \frac{d}{dy}\left[x'\right] = -\frac{d}{dy}\left[\frac{1}{4}yx^{-1}\right]
    x'' = -\frac{1}{4}x^{-1} + \frac{1}{4}yx^{-2} x'
    Now we substitute the formula for x'
    x'' = -\frac{1}{4}x^{-1} + \frac{1}{4}yx^{-2} \left(-\frac{y}{4x}\right)
    x'' = -\frac{1}{4}x^{-1} - \frac{y^2}{16x^3} \Rightarrow x'' = \frac{-4x^2 - y^2}{16x^3}.
    Last edited by lvleph; October 17th 2010 at 03:35 PM.
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  5. #5
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    Great explanation.
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