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calculus PLEASE help !

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find the equation for the line tangent to the curve at the given point.
y=((x^2)+1)/((x^2)-1) at (2, 5/3)

  • calculus PLEASE help ! -

    if you got

    y' = -4x/(x^2+1)^2

    you should be ok. That will give you a point and a slope.

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