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Partial differentiation find df/dx and df/dy

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data
    For the function of two variables f(x,y)=tan^-1(y/x)
    find df/dx and df/dy

    I know i just differentiate with respect to x and then to y but I'm stuck on the tan^-1(y/x)
    I know tan^-1(x)=1/1+X^2 when I applied this with respect to x I get 1/-1+y
    I think this is wrong please help!
  2. jcsd
  3. Nov 14, 2008 #2
    You should use the chain rule

    \frac{\partial}{\partial x}\tan^{-1}\left(\frac{y}{x}\right)=\left.\frac{d}{du}\tan^{-1}(u)\right|_{u=\frac{y}{x}}\frac{\partial}{\partial x}\frac{y}{x}
  4. Nov 14, 2008 #3


    Staff: Mentor

    That's not true.
    What is true is that
    [tex]\frac{d}{dx} tan^{-1}(x) = \frac{1}{1 + x^2}[/tex]
  5. Nov 14, 2008 #4


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    Science Advisor

    The derivative of tan-1(u) is
    [itex]\frac{1}{1+ u^2}[/itex]
    with u= y/x, that is
    [itex]\frac{1}{1+ \frac{y^2}{x^2}}[/itex]
    for the derivative with respect to x or y those are multiplied by the derivative of y/x with respect to x and the derivative of y/x with respect to x "respectively".
  6. Nov 14, 2008 #5
    so df/dx = -2x/1+y^2 ?
  7. Nov 14, 2008 #6


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    Homework Helper

    How did you get that? Halls told you it's 1/(1+y^2/x^2) times the x derivative of y/x. What's the x derivative of y/x?
  8. Nov 14, 2008 #7
    Sorry just saw what I did there. Oops. the x derivative of y/x is -y/x^2. Sorry confused myself there. Thanks for the help everyone!
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