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Math Help - Derivatives of inverse trig function

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    Derivatives of inverse trig function

    I need help taking the derivative of y = arctan(x+1)
    This is what I do
    y = arctan(x+1)
    x=tan(y+1)
    cos^2(y+1) = y'
    cos^2(arctan(x+1)+1) = y'
    cos^2(((1-sec^2x)^0.5)+2))=y'

    I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help?
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  2. #2
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    Re: Derivatives of inverse trig function

    Quote Originally Posted by Barthayn View Post
    I need help taking the derivative of y = arctan(x+1)
    This is what I do
    y = arctan(x+1)
    x=tan(y+1)
    cos^2(y+1) = y'
    cos^2(arctan(x+1)+1) = y'
    cos^2(((1-sec^2x)^0.5)+2))=y'

    I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help?
    The tan and arctan functions don't work that way. You can only take the tangent of the whole expression

    \tan(y) = \tan(\arctan(x+1))

    \tan(y) = x+1

    x = \tan(y) - 1

    When you differentiate wrt x you should get: 1 = \sec^2(y)y'
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