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Math Help - deriviation

  1. #1
    Member srirahulan's Avatar
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    Arrow deriviation

    \frac{d(\tan^{-1}\left(\frac{2x}{1-x^2}\right)}{d\tan^{-1}x},In this case how can i do this , can i use any substitution or any other method?????
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  2. #2
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    Re: deriviation

    Quote Originally Posted by srirahulan View Post
    \frac{d(\tan^{-1}\left(\frac{2x}{1-x^2}\right)}{d\tan^{-1}x},In this case how can i do this , can i use any substitution or any other method?????
    I won't swear to this but here goes

    $v(x)=\tan^{-1}\left(\dfrac {2x}{1-x^2}\right)$

    $u(x)=\tan^{-1}(x)$

    $x(u)=\tan(u)$

    $\dfrac {dv}{du} = \dfrac {dv}{dx} \dfrac{dx}{du}$

    $\dfrac{dv}{dx}=\dfrac{2}{1+x^2}$

    $\dfrac{dx}{du}=\sec^2(u)=\sec^2(\tan^{-1}(x))=1+x^2$

    $\dfrac {dv}{du} = \dfrac {dv}{dx} \dfrac{dx}{du}=\dfrac{2}{1+x^2}\cdot(1+x^2)=2$

    so

    $\dfrac{d\left(\tan^{-1}\left(\dfrac {2x}{1-x^2}\right)\right)}{d\left(\tan^{-1}(x)\right)}=2$
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  3. #3
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    Re: deriviation

    Doing a quick parametric plot I see that this is correct.
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