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)
if you got
y' = -4x/(x^2+1)^2
you should be ok. That will give you a point and a slope.
y=((x^2)+1)/((x^2)-1) at (2, 5/3)
y' = -4x/(x^2+1)^2
you should be ok. That will give you a point and a slope.