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Thread: inverse trig

  1. #1
    Newbie Wattssan's Avatar
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    inverse trig

    1.a given that y =tan-1((1+x)/(1-x)) and x<>1, show that y' = 1/(1+x2)
    b. hence given that x<1, show that tan-1((1+x)/(1-x)) -tan-1=Pi/4
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  2. #2
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    Re: inverse trig

    Quote Originally Posted by Wattssan View Post
    1.a given that y =tan-1((1+x)/(1-x)) and x<>1, show that y' = 1/(1+x2)
    b. hence given that x<1, show that tan-1((1+x)/(1-x)) -tan-1=Pi/4
    You are missing something again.

    Anyway, for derivatives of inverse trig functions, you have two choices. Memorize the formula, or use implicit differentiation:

    $y = \tan^{-1}\left(\dfrac{1+x}{1-x}\right) \Longrightarrow \tan y = (1+x)(1-x)^{-1}$

    $\sec^2 y \dfrac{dy}{dx} = (1-x)^{-2}(1+x)+(1-x)^{-1} = \dfrac{2}{(1-x)^2}$

    If $\tan y = \dfrac{1+x}{1-x} = \dfrac{\text{opp}}{\text{adj}}$, then $\cos y = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{1-x}{\sqrt{2+2x^2}}$, so:

    $\dfrac{dy}{dx} = \dfrac{2(1-x)^2}{(1-x)^2(2+2x^2)} = \dfrac{1}{1+x^2}$

    b) You do not have a complete problem.
    Thanks from topsquark
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