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Partial Differentiation: second partial derivative

  1. Mar 31, 2014 #1
    I am not quite sure how [tex]\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right)[/tex]
    [tex]=\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)[/tex]

    comes to [tex]\frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)[/tex]...


    I would have evaluated this to be the same but without the [itex]\frac{\partial z}{\partial x}[/itex] term. i.e.:

    [tex]u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial y}\right)[/tex]

    I know I am probably overlooking something obvious/simple but I can't see what it is. Could anyone tell me?
     
  2. jcsd
  3. Mar 31, 2014 #2

    Mark44

    Staff: Mentor

    How are z, u, and v related? I can guess, but that really is information you should supply.
     
  4. Mar 31, 2014 #3
    Sorry. [itex]z =f (x,y)[/itex], with

    [tex]x=\frac{1}{2}(u^2-v^2)[/tex] and [itex] y =uv[/itex]
     
  5. Mar 31, 2014 #4

    Mark44

    Staff: Mentor

    Do you understand this part?
    [tex]\frac{\partial z}{\partial u}=\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right)[/tex]

    Or are you having difficulties with this part?
    [tex] \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} )[/tex]

    [tex]= \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial z}{\partial x}\right)+v \frac{\partial}{\partial u} \left( \frac{\partial z}{\partial y} \right)[/tex]
     
  6. Mar 31, 2014 #5
    This is the part that I don't understand. The first part is OK...
     
  7. Mar 31, 2014 #6

    Mark44

    Staff: Mentor

    $$ \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} ) $$
    $$ = \frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) + \frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )$$

    Now use the product rule on each part.
     
  8. Apr 1, 2014 #7
    Thanks a lot, Mark... I am getting closer but have one problem. Having done it several times over, the same thing keeps happening:

    Using the product rule [itex](uv)' = u'v + uv'[/itex] I get one extra term.

    I'll type it piece by piece for clarity:


    [tex]\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x}) = (uv)'[/tex]

    Gives:

    [tex]\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x}) [/tex]


    And


    [tex]\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} ) = (uv)'[/tex]

    Gives:

    [tex]\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )[/tex]


    So, in total I get:

    [tex]\frac{\partial}{\partial u}( u \frac{\partial z}{\partial x})+\frac{\partial}{\partial u}(v\frac{\partial z}{\partial y} )[/tex]

    [tex]=\frac{\partial z}{\partial x}+u\frac{\partial}{\partial u}( \frac{\partial z}{\partial x})+\frac{\partial v}{\partial u}\cdot \frac{\partial z}{\partial y}+v\frac{\partial}{\partial u}(\frac{\partial z}{\partial y} )[/tex]
     
  9. Apr 1, 2014 #8

    Mark44

    Staff: Mentor

    You have z = f(x, y), where x = g(u, v) and y = h(u, v). Here x and y are dependent variables (depending on u and v), and u and v are independent variables. Since u and v are independent (are unrelated to each other), ## \frac{\partial v}{\partial u} = 0##, so your third term vanishes. By the same reasoning, ## \frac{\partial u}{\partial v}## would also be zero.

    Nice job on the LaTeX formatting!
     
  10. Apr 1, 2014 #9
    Oh, I see now. Thanks a lot, it all makes perfect sense now!
     
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