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The partial derivatives of arctan(y/x)

  1. Jan 31, 2008 #1
    [SOLVED] The partial derivatives of arctan(y/x)

    let w = arctan(y/x)

    the partial derivatives are:
    dw/dx and dw/dy

    i know that the derivative or arctan(x) is 1/(1+x^2).
    so for dw/dy, i get (1/ 1 + (y^2/x^2) ) * (1/x) = x/(x^2 + y^2) ??? correct?

    how do i find dw/dx?
     
  2. jcsd
  3. Jan 31, 2008 #2
    That looks right. To get the other, just take d/du(arctan(u)) * du/dx, where u = y/x, just like in the previous situation.
     
  4. Mar 24, 2008 #3
    The partials are:

    dw/dy = 1/ (x + y^2/x )

    &

    dw/dx = y/( 1+(y^2/x^2)

    The total integral, then, is as follows.

    w = arctan(y/x)
    dw= d(arctan(y/x) d(y/x)
    dw= 1/(1+(y/x)^2) (ydx-xdy)/(x^2)
    dw= (ydx-xdy)/[x^2(1+(y/x)^2)]
    dw= (ydx-xdy)/(x^2+y^2)
     
  5. Jan 14, 2010 #4
    Re: [SOLVED] The partial derivatives of arctan(y/x)

    although it has passed a lot of time, your answer has helped me i thank you for it, but i have something to say: i think you have written something in the uncorrect order, that's 'dw= (ydx-xdy)/(x^2+y^2)' , because if the function is y/x, the rule of the chain would be: dy*x-dx*y, then the final result (from my point of view) is:

    dw= (dy*x-x*dy)/(x^2+y^2) =-y/(x^2+y^2).

    please if i'm wrong, simply tell me.
     
  6. Jan 14, 2010 #5
    Re: [SOLVED] The partial derivatives of arctan(y/x)

    please check my answer above
     
  7. Dec 23, 2010 #6
    Re: [SOLVED] The partial derivatives of arctan(y/x)

    And what would be the time derivative of atan(y/x), where y and x are both functions of time. In other words, what is d(theta)/dt, the time derivative of the spherical coordinate theta?
     
  8. Aug 28, 2011 #7
    Re: [SOLVED] The partial derivatives of arctan(y/x)

    Fisrt, let the fucntion be W =arctan(y/x)
    Also, let W=arctan(U), such that U=y/x,
    then,
    dW= (dW/dU)(dU/dx) + (dW/dU)(dU/dy).........1
    But,
    dW/dU= 1/(1+U**2)...........................................2
    Also,
    dU/dx= -y(x)**-2, and

    dU/dy= 1/x

    .: dW= (1/(1+U**2))(-y(x)**-2) + (1/(1+U**2))(1/x)..........3

    Putting the values U in Equation 3,

    dW= (1/(1+(y/x)**2)(-y(x)**-2 + (1/(1+(y/x)**2)(1/x)

    = (1/(1+(y/x)**2)((-y/x) + (1/x)) or

    ((1/(1+(y/x)**2)((x-y)/(x)**2)).
     
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