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Math Help - I cannot solve this...More partial derivatives...

  1. #1
    Junior Member
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    I cannot solve this...More partial derivatives...

    I have read the book, tried looking at similar examples, but I simply cannot solve the following problem:

    The function f(u,v) is differentiable in the entire R^2. Write:

    h(x,y,z)= f(x/y, y/z)

    y>0, z> 0
    (do not understand why this is of importance)

    Calculate:

    x*(dh/dx)+y*(dh/dy)+z*(dh/dz) expressed in u, v and partial derivates of f.

    The dh/dx, dh/dy and dh/dz all have those crooked d's which I cannot type here.

    I really need help with this and would be so thankful for any guidance!
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  2. #2
    Senior Member
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    \frac {\partial h}{\partial x} \: = \frac {\partial f}{\partial u} \frac {\partial u}{\partial x} + \frac {\partial f}{\partial v} \frac {\partial v}{\partial x} =\frac {\partial f}{\partial u}\frac {1}{y}

    where

    \frac {\partial u}{\partial x} =\frac {\partial (x/y)}{\partial x} =1/y

    \frac {\partial v}{\partial x} =\frac {\partial (y/z)}{\partial x} =0

    As

    u=x/y \: it  \: follows  \: that \: y \ne 0
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  3. #3
    Junior Member
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    Thanks, but should I do the calculation for dh/dy and dh/dz as well...?

    I get:

    dh/dy= (df/du)*(-x/y^2)+(df/dv)*(1/z)
    dh/dz= (df/dv)*(-y/2)

    Shall I mulitply this with x, y and z respectively? The key answer says:
    x*(dh/dx)+y*(dh/dy)+z*(dh/dz)= 0
    But I dont understand how...

    Help would be gold...
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  4. #4
    Senior Member
    Joined
    Mar 2010
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    <br />
x\frac {\partial h}{\partial x}=\frac {x}{y}\frac {\partial f}{\partial u}<br />

     y\frac {\partial h}{\partial y}=...

     z\frac {\partial h}{\partial z}=...

    and make sum of three terms.
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