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Partial Derivatives

  1. Jun 22, 2009 #1
    Hey everybody, first time poster although i've recently come across this forum and it's helped me discover the solution of many problems ive been having. I've seen to come to grips with most partial derivative problems i've come across, however, i still cant get correct solutions to problems involving quotients, dont know how exactly to go about it. The two specific questions ive got in my book which i cant seem to get are given below.

    1. The problem statement, all variables and given/known data

    Evaluate the first partial derivative

    2. Relevant equations

    f(x, y) = (x + y)/(x - y) (1)

    f(x, y) = log(1+x) / log(1+y) (2)

    Thanks for any help which could be given =)

    I assume you use the standard quotient rule, but i dont know how exactly to go about it with multiple variables...thanks.
  2. jcsd
  3. Jun 22, 2009 #2

    Doc Al

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    Staff: Mentor

    Yes, you'd use the quotient (or product) rule and the chain rule. When computing ∂F/∂x, treat y as a constant; When computing ∂F/∂y, treat x as a constant.
  4. Jun 22, 2009 #3


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    Gold Member

    Yes, you simply use the quotient rule, keeping the variable you're not taking the partial derivative of as a constant.

    f(x,y) = \frac{{g(x,y)}}{{h(x,y)}} \\
    \frac{{\partial f(x,y)}}{x} = \frac{{\frac{{\partial g(x,y)}}{{\partial x}}h(x,y) - g(x,y)\frac{{\partial h(x,y)}}{{dx}}}}{{h(x,y)^2 }} \\

    is an example of the partial with respect to x where you already know how to properly take partial derivatives of functions.
  5. Jun 22, 2009 #4
    Thanks for the help guys but im still getting something wrong, which im guessing is a small but stupid mistake...

    for df/dx i get 1/(1+x) . -1(log(1+y))^-2 (unless thats right?)

    for df/dy im confused...whats the derivative of a log of a constant :S

    Thanks for any help =)
  6. Jun 22, 2009 #5
    Well the derivative of a constant is 0. And the log of a constant is just another number.
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