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Math Help - Partial Derivative and chain rule...

  1. #1
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    Partial Derivative and chain rule...

    Hi guys I am having difficulty while revising..

    Consider the function r=(x^2+y^2+z^2) and f=r^-1

    Find Partial derivatives rx,ry,rz. I have done this part but having difficulty doing this part

    Use chain rule to find fx,fy,fz and hence show that grad ∆f =(-r^-3)(r) where r=xi+yj+zk


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  2. #2
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    Here, the Chain Rule states

    \frac{\partial f}{\partial x}=\frac{df}{dr}\frac{\partial r}{\partial x}.

    We have, by calculation

    \frac{df}{dr} = -\frac{1}{r^2}\;\;\;\;\;\;\;\;\;\;\frac{\partial r}{\partial x} = 2x.<br />

    Thus, we have

    \frac{\partial f}{\partial x}=\left(-\frac{1}{r^2}\right)\cdot(2x)=-\frac{2x}{r^2}.

    I'm not sure what you mean by \mbox{grad}\,\Delta f, but for \nabla f I calculated something that looked quite different from the answer given in the problem.
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  3. #3
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    Quote Originally Posted by Scott H View Post
    Here, the Chain Rule states

    \frac{\partial f}{\partial x}=\frac{df}{dr}\frac{\partial r}{\partial x}.

    We have, by calculation

    \frac{df}{dr} = -\frac{1}{r^2}\;\;\;\;\;\;\;\;\;\;\frac{\partial r}{\partial x} = 2x.<br />

    Thus, we have

    \frac{\partial f}{\partial x}=\left(-\frac{1}{r^2}\right)\cdot(2x)=-\frac{2x}{r^2}.

    I'm not sure what you mean by \mbox{grad}\,\Delta f, but for \nabla f I calculated something that looked quite different from the answer given in the problem.



    I see thanx...actually grad f is same as saying \nabla f I didnt know how to use the symbol...
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  4. #4
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    oo the function is actually supposed to be r=(x^2+y^2+z^2)^1/2 that is probably why you didnt get the same answer as the one I posted...sorry I missed that..and thanx
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