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Partial Derivatives

  1. Dec 10, 2007 #1
    [SOLVED] Partial Derivatives

    I'm having a bit of trouble on an old test problem. It states:

    Determine if there is a function f(x, y) such that fx(x, y) = yex + 1 and fy(x, y) = ex + cos(y). If such a function exists, find it.

    I know that such a function exists because fxy(x, y) = ex, and fyx(x, y) = ex, thus fxy = fyx.

    I'm having trouble finding the original function. I know that I would probably have to integrate somehow, but I'm not sure how to go about that with partial derivatives.

    Any help is appreciated.
  2. jcsd
  3. Dec 10, 2007 #2


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    integrate fx(x, y) with respect to x. This will tell you that f(x,y) is of the form e^x+x+C(y), where C(y) is some function of y. Do the same thing, but with fy(x, y).. you'll get f(x,y)=...+D(x) where CDx) is some function of x. Compare the two expressions of f(x,y) thus obtained, and conclude as to the value of C(y) and D(x), and hence of f(x,y)
  4. Dec 10, 2007 #3
    Ok, I found that f(x, y) = yex + x + sin(y), and it checks with the first order partial derivatives. Thank you very much :)
  5. Dec 11, 2007 #4


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    Plus a constant, of course.
  6. Dec 11, 2007 #5
    Right, I forgot that part, haha. Thanks :)
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