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Math Help - Partial differentiation

  1. #1
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    Second derivative of partial differentiation???!!

    Consider the function of two variables

    u(x,y) = f(x-y) + g(x+y/3)

    where f(s) and g(t) are each arbitrary functions of a single variable.

    Using the change of variables

    s=x-y and t=x+y/3

    what are the first and second derivatives of u with respect to x and y in terms of f and g??thankssss
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by bryan06 View Post
    Consider the function of two variables

    u(x,y) = f(x-y) + g(x+y/3)

    where f(s) and g(t) are each arbitrary functions of a single variable.

    Using the change of variables

    s=x-y and t=x+y/3

    what are the first and second derivatives of u with respect to x and y in terms of f and g??thankssss
    Use the chain rule

    \frac{\partial u}{\partial x}=\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial x}

    Now we need to take the 2nd derivative

    \frac{\partial }{\partial x}\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}\right)+\frac{\partial }{\partial x}\left( \frac{\partial f}{\partial t}\frac{\partial t}{\partial x} \right)

    Now we need to use the product rule (and chain) on both of the above terms to get


    \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 f}{\partial s^2}\left(\frac{\partial s}{\partial x}\right)^2+\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial x^2}+\frac{\partial^2 f}{\partial t^2}\left(\frac{\partial t}{\partial x}\right)^2+\frac{\partial f}{\partial t}\frac{\partial^2 t}{\partial x^2}

    See if you can finish
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  3. #3
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    thanksss~~~yess i did manage to finish it its very kind of you indeed!

    Quote Originally Posted by TheEmptySet View Post
    Use the chain rule

    \frac{\partial u}{\partial x}=\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial x}

    Now we need to take the 2nd derivative

    \frac{\partial }{\partial x}\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial x}\right)+\frac{\partial }{\partial x}\left( \frac{\partial f}{\partial t}\frac{\partial t}{\partial x} \right)

    Now we need to use the product rule (and chain) on both of the above terms to get


    \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 f}{\partial s^2}\left(\frac{\partial s}{\partial x}\right)^2+\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial x^2}+\frac{\partial^2 f}{\partial t^2}\left(\frac{\partial t}{\partial x}\right)^2+\frac{\partial f}{\partial t}\frac{\partial^2 t}{\partial x^2}

    See if you can finish
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