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Math Help - Second Partial Differentiation - Explain this to me please!

  1. #1
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    Second Partial Differentiation - Explain this to me please!

    I need help with the explanation for this one.

    If T (r)

    and r = \sqrt{x^2+ y^2}


    This makes

    \frac{\partial T}{\partial x} = \frac{x}{r}\frac{dT}{dr}

    Now I need to find the second derivative:

    Using operators

    \frac{\partial}{\partial x}(T) = \frac{x}{r}\frac{d}{dr}(T)

    For the second derivative it would become

    \frac{\partial}{\partial x}(\frac{dT}{dx}) = \frac{\partial}{\partial x}(\frac{x}{r}\frac{dT}{dr})

    I understand upto this point, but then I do not understand how this turns up becoming:

    \frac{x}{r}\frac{\partial}{\partial r}(\frac{x}{r}\frac{dT}{dr})

    Can somebody please tell me what happened in between??


    Thank you very much!!
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  2. #2
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    Quote Originally Posted by gva0324 View Post
    I need help with the explanation for this one.

    If T (r)

    and r = \sqrt{x^2+ y^2}


    This makes

    \frac{\partial T}{\partial x} = \frac{x}{r}\frac{dT}{dr}

    Now I need to find the second derivative:

    Using operators

    \frac{\partial}{\partial x}(T) = \frac{x}{r}\frac{d}{dr}(T)

    For the second derivative it would become

    \frac{\partial}{\partial x}(\frac{dT}{dx}) = \frac{\partial}{\partial x}(\frac{x}{r}\frac{dT}{dr})

    I understand upto this point, but then I do not understand how this turns up becoming:

    \frac{x}{r}\frac{\partial}{\partial r}(\frac{x}{r}\frac{dT}{dr})

    Can somebody please tell me what happened in between??


    Thank you very much!!
    It might be easier to see the general chain rule

     <br />
T_x = T_r \cdot r_x<br />

     <br />
T_{xx} = T_{rr} \cdot \left(r_x\right)^2 + T_r \cdot r_{xx}.<br />
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