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Homework Help: Partial derivatives

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi there. Well, I got the next function, and I'm trying to work with it. I wanted to know if this is right, I think it isn't, so I wanted your opinion on this which is always helpful.

    [tex]f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ si }& y\neq-x\\0 & \mbox{si}& y=-x\end{matrix}[/tex]

    If [tex]y\neq-x[/tex]:
    [tex]\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

    [tex]\displaystyle\frac{\partial f}{\partial y}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

    Second case:

    If [tex]y=-x[/tex]

    [tex]\displaystyle\frac{\partial f}{\partial y}=\displaystyle\frac{\partial f}{\partial x}=0[/tex]

    Is this right? I would like to know why it isn't, I think its not. Should I think when I consider the second case as I would go in all directions or something like that? cause in other cases where I got defined different functions for a particular point I've used the definition, but in this case I got trajectories. What should I do?

    Bye and thanks.
  2. jcsd
  3. Sep 18, 2010 #2


    Staff: Mentor

    You're not using the product rule correctly in either partial. With ordinary derivative, if h(x) = f(x)*g(x), h'(x) = f'(x)*g(x) + f(x)*g'(x). It's similar for your two partials.

    Also, when you differentiate sin(pi/(x + y)), you're not using the chain rule correctly. You will get cos(pi/(x + y)), but now you need the derivative of the argument to the cosine function, so you need the derivative of pi/(x + y).
  4. Sep 19, 2010 #3
    Why not?


    [tex]\displaystyle\frac{\partial f}{\partial x}=2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{(x+y)})[/tex]

    The derivative on the left is obvious. For the right part, I've need to find derivative of [tex]\sin(\displaystyle\frac{\pi}{x+y})[/tex], which is the derivative of the sine multiplied by the derivative on the inside of the sine (Im working with respect to x):
    Then, when I make the product with [tex](x+y)^2[/tex] I got [tex]-\pi\cos(\displaystyle\frac{\pi}{x+y})[/tex]

    I don't see whats wrong with it.
  5. Sep 19, 2010 #4


    Staff: Mentor

    You're right. I didn't work the problems through, but it seemed that you were missing a factor on the second terms of your partials. Your work didn't show that intermediate step (before canceling), so I assumed you had not used the chain rule correctly.
  6. Sep 19, 2010 #5
    Thanks. Why I have to use the definition of derivative in the second case?
  7. Sep 19, 2010 #6


    Staff: Mentor

    No, I don't see why you would need to use the definition of the derivative. Since f(x, y) = 0 along the line y = -x, both partials are zero as well.
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