y= 2x/ x^2 +1 Find an equation of the tangent line to the curve at (1,1)

Answer 1

First need to determine the slope of the function which is the derivative of the function so lets find the derivative. To do this need to use the quotient rule so y’( 2x/x^2+1)=

[(2)(x^2+1) - (2x)(2x)/ (x^2+1)^2

2x^2 +2 -4x^2/(x^2+1)^2= -2x^2 +2/(x^2+1)^2 So y’(1)= -2(1)+2/(1^2+1)^2= 0

So slope at x=1 is 0

To find the equation of the line at x=1 use y=mx+ b to solve for b

1=0(1) +b

b=1

y=(0)(1) +1

y=1 is the equation of the tangent line

Answer 2

y = 2x / (x² + 1)

dy/dx = [2(x² + 1) - 4x²] / (x² + 1)²

Slope of tangent line at (1,1) is dy/dx evaluated at x = 1.

So, slope = 0. Therefore the tangent line is horizontal at (1,1).

Equation of tangent line is y = 1