1. Limited time only! Sign up for a free 30min personal 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!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook