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Partial derivatives

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    In the steps below, the ∂z/∂x does not seem to be obeying normal algebraic rules. I'm confused. This is not really a problem, I'm just trying to understand the steps.

    3. The attempt at a solution

    1. 3z2∂z/∂x - y + y∂z/∂x = 0
    2. ∂z/∂x = y/(y + 3z2)

    if ∂z/∂x were a normal algebraic variable, say p, it would be

    2p = y/(y + 3z2)

    so why not

    2∂z/∂x = y/(y + 3z2)
     
  2. jcsd
  3. Mar 4, 2012 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    I don't see how you got that extra factor of 2.
    \begin{align}&3z^2 p - y + yp = 0\\
    &(3z^2+y)p = y\\
    &p =\frac{y}{3z^2+y}
    \end{align}
     
  4. Mar 4, 2012 #3

    Mark44

    Staff: Mentor

    bobsmith,
    What Fredrik did in his 2nd step was to factor out p (AKA ∂z/∂x), essentially doing the opposite of the distributive property. This is yet another example of where weakness in your algebraic knowledge is preventing you from understanding very simple operations. Until and unless you rectify this problem, you will continue to be unable to make sense of what you're reading.
     
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