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Partial differentiaton question, how to distinguish between u and f (u )

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Partial differentiation question

    V=xf(u) and u=y/x, show that x^2. d2v/dx2 + 2xy. d2v/dxdy + y^2. d2V/dy2 = 0
    Is u the same as f(u)

    2. Relevant equations

    V=xf(u) and u=y/x

    3. The attempt at a solution
    do you just differentiate the product of x.y/x
     
  2. jcsd
  3. Aug 23, 2011 #2

    vela

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    Science Advisor
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    No, u is not the same as f(u).

    You need to use the product and chain rules. For example, to calculate ∂V/∂x, you'd start by applying the product rule to get
    \begin{equation*}
    \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}[xf(u)] = f(u) + x \frac{\partial f(u)}{\partial x}
    \end{equation*}
    To calculate the second term, you need to use the chain rule, which says
    \begin{equation*}
    \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} = -\frac{y}{x^2}\frac{\partial f}{\partial u}
    \end{equation*}
    So you have
    \begin{equation*}
    \frac{\partial V}{\partial x} = f(u) - \frac{y}{x} \frac{\partial f}{\partial u} = f(u) - u\frac{\partial f}{\partial u}
    \end{equation*}
     
    Last edited: Aug 23, 2011
  4. Aug 23, 2011 #3
    Thankyou very much but i still cannot understand how that may lead to the form which the question wants us to show, thanks.
     
  5. Aug 23, 2011 #4

    Mark44

    Staff: Mentor

    Well, you have a way to go. Starting from vela's advice, you need to calculate ∂2V/∂x2, ∂V/∂y, ∂2V/∂y2, and ∂2V/(∂x∂y), and then substitute them into the left side of your equation, which should simplify to zero.
     
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