1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with this partial derivative problem

  1. Jun 8, 2005 #1
    I'm supposed to find (assume all these d's are the partial derivative sign, not d)

    d^2z/dxdy, d^2z/dx^2, and d^2z/dy^2

    The one I can't do is z^2 + sinx = tany

    I set it equal to zero, so z^2 + sinx - tany=0

    so dz/dx = - Fx/Fz = sec^2y/2z
    dz/dy = - Fy/Fz = -cosx/2z

    multiply them for d^2z/dxdy I get -(sec^2y*cosx)/(4z^2) which is ALMOST right, but the book says the denominator is 4z^3, and the other two I do the same procedure(dz/dx * dz/dx for the second one for example) and get nothing close
     
  2. jcsd
  3. Jun 8, 2005 #2
    Your problem seems to be the fact that
    [tex]\frac{\partial^2z}{\partial x\partial y} \neq \frac{\partial z}{\partial x}\frac{\partial z}{\partial y}.[/tex]

    On the left-hand side (what you want to find), you are taking the partial derivative with respect to x of the partial derivative of z with respect to y, that is
    [tex]\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right).[/tex]
    This gives you a second-order derivative, whereas on the right-hand side (what you did) you are merely multiplying two first-order derivatives together... The same will be true for the other derivatives you have to find, since
    [tex]\frac{\partial^2z}{\partial x^2} \neq \left(\frac{\partial z}{\partial x}\right)^2.[/tex]
     
  4. Jun 8, 2005 #3
    Ah, ok, one mistake outta the way

    in that case, and I know I'm doing it wrong, if I take the partial derivative of z with respect to y, I end up with 2z*dz/dy + sec^2y, and if I take the partial derivative with respect to x at this point, I end up with 0, which clearly isn't right. I can usually do implicit differentiation but I'm doing something silly here

    Edit, oh wait I'm stupid, I guess the -Fx/Fy stuff was still right then? Hang on

    Edit #2 - Oh jeez, I had that bacwards up there too, so I'm still taking the partial with respect to x, but with no x in the equation which gives me 0:(
     
    Last edited: Jun 8, 2005
  5. Jun 8, 2005 #4
    YES, got it, though I'm not sure it was the most efficient way
    I set the equation equal to z to find dz/dy,
    z=sqrt(tanx-sinx)
    dz/dy=-((tany-sinx)^-1/2 / 2) * sec^2y

    then derivative in terms of x, (-sec^2 y /2) * 1/2(tany-sinx)^-3/2 * cosx

    plug in z for (tanx-sinx)^1/2 and that all simplifies to the right answer
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with this partial derivative problem
  1. Partial Derivation (Replies: 7)

Loading...