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Thread: A 'trivial' question...

  1. #1
    MHF Contributor chisigma's Avatar
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    A 'trivial' question...

    It is well known that the derivative of a 'function of function' y\{u(x)\} is...

    \frac {dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

    A question for you: does it exist a similar expression for the second derivative \frac {d^{2} y}{dx^{2}}?...

    Any help will be greatly appreciated!...

    Kind regards

    \chi \sigma
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  2. #2
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    Quote Originally Posted by chisigma View Post
    A question for you: does it exist a similar expression for the second derivative \frac {d^{2} y}{dx^{2}}?...
    The answer is usually expressed in slightly different terms.
    If y=f(u) and u=g(x) where f~\&~g are both twice differentiable then
     \frac{{d^2 y}}{{dx^2 }} = \frac{{d^2 y}}{{du^2 }}\left( {\frac{{du}}{{dx}}} \right)^2  + \frac{{dy}}{{du}}\frac{{d^2 u}}{{dx^2 }}.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by chisigma View Post
    It is well known that the derivative of a 'function of function' y\{u(x)\} is...

    \frac {dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

    A question for you: does it exist a similar expression for the second derivative \frac {d^{2} y}{dx^{2}}?...

    Any help will be greatly appreciated!...

    Kind regards

    \chi \sigma

    \left(\frac{d^2 y}{du^2}\cdot \frac{du}{dx}\right)\left(\frac{du}{dx}\right)+\le  ft(\frac{d^2 u}{dx^2}\right)\left(\frac{dy}{du}\right)

    Edited:- I'm late , Plato beat me .
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  4. #4
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    Quote Originally Posted by Plato View Post
    The answer is usually expressed in slightly different terms.
    If y=f(u) and u=g(x) where f~\&~g are both twice differentiable then
     \frac{{d^2 y}}{{dx^2 }} = \frac{{d^2 y}}{{du^2 }}\left( {\frac{{du}}{{dx}}} \right)^2  + \frac{{dy}}{{du}}\frac{{d^2 u}}{{dx^2 }}.
    The reason is that \frac{d^2y}{dx^2} = \frac d{dx}\Bigl(\frac{dy}{dx}\Bigr) = \frac d{dx}\Bigl(\frac{dy}{du}\frac{du}{dx}\Bigr) = \frac d{dx}\Bigl(\frac{dy}{du}\Bigr)\frac{du}{dx} + \frac{dy}{du}\frac{d^2u}{dx^2} (product rule). But

    \frac d{dx}\Bigl(\frac{dy}{du}\Bigr) = \frac d{du}\Bigl(\frac{dy}{du}\Bigr)\frac{du}{dx} = \frac{d^2y}{du^2}\frac{du}{dx}.
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  5. #5
    MHF Contributor chisigma's Avatar
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    For you all: thanks very much! ...

    ... the purpose of my question is related to the following differential equation...

     x^{2} \cdot y^{''} + x\cdot y^{'} + y = \sin (\ln x) (1)

    ... that has been proposed in another section...

    A 'brute force approach' to (1) seems to have little chances but if we operate the change of variables u= \ln x , thanks to the formula you have supplied, we have...

    \frac {dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} = \frac{1}{x}\cdot \frac{dy}{du}

    \frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{du^{2}}\cdot (\frac{du}{dx})^{2} + \frac{dy}{du}\cdot \frac{d^{2}u}{dx^{2}} = \frac{1}{x^{2}}\cdot (\frac{d^{2}y}{du^{2}} - \frac{dy}{du}) (2)

    ... and the (1) becomes...

     y^{''} + y = \sin u (3)

    ... that is much more tractable [linear DE with constant coefficients...] and has as solution...

     y = c_{1} \cos u + c_{2} \sin u + \frac{u}{2}\cdot \cos u (4)

    thanks again! ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; Aug 22nd 2009 at 12:32 AM. Reason: error in (4)... sorry!...
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