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Thread: Implicit differentiation.

  1. #1
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    Post Implicit differentiation.

    Hi All!

    Having a problem when differentiating for the second time $\displaystyle \frac{d^2y}{dx^2}$an implicit equation.

    I'll give two example problems:

    $\displaystyle 2x^2 + 2y^2 - 4x + 6 = 0}$
    $\displaystyle = 4x + 4y.\frac{dy}{dx} -4 = 0$
    $\displaystyle \frac{dy}{dx} = \frac{4 - 4x}{4y}$

    Not to sure how to take it from here to do the second derivative $\displaystyle (\frac{d^2y}{dx^2})$

    Second problem:

    $\displaystyle y^2 + x^2 = 1 $
    $\displaystyle 2y.\frac{dy}{dx} + 2x = 0 $
    $\displaystyle \frac{dy}{dx} = -\frac{2x}{2y}$

    Also not sure how to take it from here to get $\displaystyle \frac{d^2y}{dx^2}$.

    Your help will be appreciated. Thanks

    Regards,

    dadon
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  2. #2
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    Quote Originally Posted by dadon

    Second problem:

    $\displaystyle y^2 + x^2 = 1 $
    $\displaystyle 2y.\frac{dy}{dx} + 2x = 0 $
    $\displaystyle \frac{dy}{dx} = -\frac{2x}{2y}$

    Also not sure how to take it from here to get $\displaystyle \frac{d^2y}{dx^2}$.

    Your help will be appreciated. Thanks

    Regards,

    dadon
    You have,
    $\displaystyle x^2+y^2=1$
    Thus,
    $\displaystyle 2x+2yy'=0$ thus, $\displaystyle y'=-\frac{x}{y}$
    Thus,
    $\displaystyle 2+2y''+2y'=0$
    Substitute, and divide by 2,
    $\displaystyle 1+y''-\frac{x}{y}=0$
    Thus,
    $\displaystyle y''=\frac{x-y}{y}$
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  3. #3
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    Sorry I don't get how you get this line:

    2 + 2y'' + 2y' = 0

    Cheers for the quick response!
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  4. #4
    TD!
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    @ ThePerfectHacker: are you sure about that? Where did your 3rd line come from?

    $\displaystyle
    \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}\left( { - \frac{x}{y}} \right) = - \frac{{y - xy'}}{{y^2 }}
    $

    Now substituting the expression we have for y' gives

    $\displaystyle
    \frac{{d^2 y}}{{dx^2 }} = \frac{{ - x\frac{x}{y} - y}}{{y^2 }} = - \frac{{x^2 + y^2 }}{{y^3 }}
    $

    First one goes the same.

    Edit: I think you were trying to take the derivative again, after 2x+2yy'=0 but don't forget that y and y' are a function of x. The derivative would give:

    $\displaystyle
    2 + 2y'^2 + 2yy'' = 0 \Leftrightarrow y'' = \frac{{ - 1 - y'^2 }}{y}
    $

    Now substituting for y' again yields the same result as above.
    Last edited by TD!; Apr 26th 2006 at 02:31 PM.
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  5. #5
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    Sorry, my fault.
    My third line should have been, (implicity again),
    $\displaystyle 2+2y'y'+2yy''=0$
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  6. #6
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    Thanks guys!

    I will do the second example and get back to you.

    Thanks agian.
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  7. #7
    TD!
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    No problem - I'll be glad to check it for you
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    Post re:

    Hey cheers!

    $\displaystyle
    2x^2 + 2y^2 - 4x + 6 = 0}
    $
    $\displaystyle
    4x + 4y.\frac{dy}{dx} -4 = 0
    $
    $\displaystyle
    \frac{dy}{dx} = \frac{4 - 4x}{4y}
    $
    $\displaystyle \frac{dy}{dx} = -\frac{x}{y}$

    This should give the same answer as before as the dy/dx are the same.

    This is what I got:

    $\displaystyle
    \frac{d^2y}{dx^2}
    $ = $\displaystyle \frac{-y^2 + x^2}{y^3}$

    Is the y meant to be negative?
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  9. #9
    TD!
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    Quote Originally Posted by dadon
    $\displaystyle
    \frac{dy}{dx} = \frac{4 - 4x}{4y}
    $
    $\displaystyle \frac{dy}{dx} = -\frac{x}{y}$

    This should give the same answer as before as the dy/dx are the same.
    What happens here? I think you get

    $\displaystyle
    \frac{{dy}}{{dx}} = \frac{{4 - 4x}}{{4y}} = \frac{{1 - x}}{y}
    $

    Now take the d/dx again, using the quotient rule and remembering that y = y(x).
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    Post re:

    oh sorry! silly mistake il try agian!
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  11. #11
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    so...

    $\displaystyle
    \frac{d^2y}{dx^2}
    $ = $\displaystyle \frac{-y^2 - (1-x)^2}{y^3}$

    Is the y meant to be negative or as it is squared just change it to positive?
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  12. #12
    TD!
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    Looks good. First we take the derivative again:

    $\displaystyle
    \left( {\frac{{1 - x}}{y}} \right)^\prime = \frac{{\left( {1 - x} \right)^\prime y - \left( {1 - x} \right)y'}}{{y^2 }} = \frac{{ - y - \left( {1 - x} \right)y'}}{{y^2 }}
    $

    Then substitute for y':

    $\displaystyle
    \frac{{ - y - \left( {1 - x} \right)y'}}{{y^2 }} = \frac{{ - y - \left( {1 - x} \right)\frac{{1 - x}}{y}}}{{y^2 }} = \frac{{ - y^2 - \left( {1 - x} \right)^2 }}{{y^3 }}
    $
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  13. #13
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    Post re:

    Thanks! Worked out nicely!

    Kind Regards,

    dadon
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  14. #14
    TD!
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    You're welcome
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