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Partial derivatites of arctan

  1. May 16, 2005 #1
    Partial derivatives of arctan

    Could anyone help me with the following partial derivatives?

    [tex]\nabla \arctan \left(\frac x y \right)[/tex]

    [tex]\nabla \arctan \left(\frac y x \right)[/tex]


    Last edited: May 16, 2005
  2. jcsd
  3. May 17, 2005 #2


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    Mmmh. Did you know that

    [tex]\frac{d}{dx}Arctg[f(x)] = \frac{f'(x)}{1+[f(x)]^2}[/tex]

  4. May 17, 2005 #3

    James R

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    Start with fact that

    [tex]\frac{d}{dx} \arctan x = \frac{1}{1+x^2}[/tex]

    then apply the chain rule.

    Does that help?
  5. May 17, 2005 #4


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    Dearly Missed

    It is of some interest to note that for any "x", we have the equality:
  6. May 17, 2005 #5
    Just to clear up notation issues, he's looking for the partial derivative for each variable right? Such as

    [tex] \nabla_x = \frac{1}{y+\frac{x^2}{y}}[/tex]
  7. May 17, 2005 #6
    For example:

    [tex]\nabla \arctan \left(\frac x y \right) = \left(\frac \partial {\partial x} \arctan \left(\frac x y \right), \frac \partial {\partial y} \arctan \left(\frac x y \right)\right)[/tex]
  8. May 17, 2005 #7
    Actually, I don't even remember how to find the derivative of [tex]\arctan(x)[/tex] (the formulas you posted). I tried something like this:

    [tex]\tan \left(\arctan(x) \right) = x[/tex]

    Differentiating in respect to [tex]x[/tex]:

    [tex]\sec ^2 \left(\arctan(x) \right).\arctan ' (x) = 1[/tex]
    [tex]\arctan '(x) = \frac 1 {\sec ^2 \left(\arctan(x) \right)}[/tex]

    Not really what I was supposed to find... Could you please give me any more hints? :uhh:
  9. May 17, 2005 #8
    Now rewrite the RHS of the last equation in terms of tan using a simple trigonometric identity.
  10. May 17, 2005 #9
    OK, I figured it out myself.

    Now I use:

    [tex]\sec^2 \left(\arctan(x) \right) = \tan^2 \left(\arctan (x) \right) + 1 = x^2 + 1[/tex]

    What gives me:

    [tex]\arctan '(x) = \frac 1 {x^2 + 1}[/tex]

    Maybe I can do something similar with the original derivatives. I'll give it a try.
  11. May 17, 2005 #10
    [tex] y = arctanx [/itex]

    [tex] tan y = x [/itex]

    Implicit differentiation wrt x:

    [tex] sec^2y \frac{dy}{dx} = 1 [/tex]

    Solve that for dy/dx. Draw a triangle, angle 'y', opposite 'x' and adjacent '1'. Find the hypotenuse, and describe the resulting expression in terms of x.

    edit: wow im slow
  12. May 17, 2005 #11
    I'm gonna try:

    [tex]\frac \partial {\partial x} \arctan \left(\frac x y \right)[/tex]

    [tex]\tan \left(\arctan \left(\frac x y\right) \right) = \frac x y[/tex]

    Derivative with respect to x:

    [tex]\sec^ 2 \left(\arctan \left(\frac x y\right) \right).\arctan' \left(\frac x y\right) . \frac 1 y = \frac 1 y[/tex]
    [tex]\arctan' \left(\frac x y\right) \right = \frac 1 {\frac {x^2} {y^2} + 1} = \frac {y^2} {x^2 + y^2}[/tex]

    Is this right?

    Thanks again
    Last edited: May 17, 2005
  13. May 17, 2005 #12
    I'm sorry thats really hard to follow, name f(x,y) = z and try going through again, for me. Your result is not the correct answer though, it looks like you forgot to apply the chain rule.
  14. May 17, 2005 #13
    Do you want me to name [tex]f (x,y) = \arctan \left(\frac x y \right) [/tex] ? I'm not sure what you mean.
  15. May 17, 2005 #14


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    Nope,it's incorrect.

    [tex] \nabla \arctan\frac{x}{y}=\frac{\partial \arctan\frac{x}{y}}{\partial x}\vec{i}+\frac{\partial \arctan\frac{x}{y}}{\partial y} \vec{j} [/tex]

    Take the first derivative

    [tex]\frac{\partial \arctan\frac{x}{y}}{\partial x}=\frac{1}{y}\frac{1}{1+\frac{x^{2}}{y^{2}}}=\frac{y}{x^{2}+y^{2}} [/tex]

    You do the second.

  16. May 17, 2005 #15
    I think I got it now:

    [tex]\frac{\partial \arctan\frac{x}{y}}{\partial y}=\frac{-x}{y^2}\frac{1}{\frac{x^{2}}{y^{2}}+1}=-\frac{x}{x^{2}+y^{2}} [/tex]

    I just can't figure out what I did wrong in the other post.

    But that's OK. Thanks a lot.
  17. May 17, 2005 #16
    You didnt apply the chain rule the first time, which you did this time. So you are fine.
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