Results 1 to 2 of 2

Thread: Partial derivatives of arccos

  1. #1
    Jun 2013

    Partial derivatives of arccos

    Hi All,

    To be able to construct a Jacobian, I am asked to find the two partial derivatives of the following inverse trigonometric function:

    w_{1}(u,v) = \frac{1}{2\pi} \arccos (\frac{u}{\sqrt{u^{2} + v^{2}}})

    I know that \frac{\delta \arccos x}{\delta x} = \frac{-1}{\sqrt{1-x^2}}, but I am realy stumped at how to proceed from this to get the partial derivatives.

    The anwers to the partian derivatives are given as:

    \frac{\delta w_{1}}{\delta u} = \frac{-v}{2\pi(u^{2}+v^{2})}

    \frac{\delta w_{1}}{\delta v} = \frac{u}{2\pi(u^{2}+v^{2})}

    Any help on how to arrive here would be much appreciated!
    Kind regards,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Apr 2005

    Re: Partial derivatives of arccos

    Use the chain rule. If w_1= \frac{1}{2\pi}arcccos(\frac{u}{\sqrt{u^2+ v^2}}, then, yes, the derivative of arcos(x) is \frac{-1}{\sqrt{1- x^2}}.
    Let x(u,v)= \frac{u}{\sqrt{u^2+ v^}} so that w_1(u)= \frac{1}{2\pi}arcos(x)
    Now, the chain rule says that \frac{\partial w_1}{\partial u}= \frac{dw_1}{dx}\frac{\partial x}{\partial u}
    You already know that \frac{dw_1}{dx}= \frac{-1}{\sqrt{1- x^2}}= \frac{-1}{\sqrt{1- \frac{u^2}{u^2+ v^2}}}
    and now you need to multiply that by \frac{\partial x}{\partial u}= \frac{\partial}{\partial u}\left(\frac{u}{\sqrt{u^2+ v^2}}\right)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. arccos(1/3)+2*arccos(-2sqrt(2)/3)
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Jun 10th 2013, 12:43 AM
  2. Replies: 2
    Last Post: Nov 12th 2012, 01:55 AM
  3. Replies: 3
    Last Post: May 4th 2012, 12:45 AM
  4. Partial Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 22nd 2011, 02:18 PM
  5. partial derivatives
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Jul 8th 2008, 10:44 AM

Search Tags

/mathhelpforum @mathhelpforum