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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:

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

    I know that $\displaystyle \frac{\delta \arccos x}{\delta x} = \frac{-1}{\sqrt{1-x^2}}$, 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:

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

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

    Any help on how to arrive here would be much appreciated!
    Kind regards,
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  2. #2
    MHF Contributor

    Apr 2005

    Re: Partial derivatives of arccos

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