Calculus: Early Transcendentals, Chapter 3, 3.2, Section 3.2, Problem 34

The slope of a tangent line to a function at a point is actually equal to the derivative of the function at that point.
y=(2x)/(x^2+1)
dy/dx=(((x^2+1)*d/dx(2x))-(2x)*(d/dx(x^2+1)))/(x^2+1)^2
dy/dx=((x^2+1)*2-2x*2x)/(x^2+1)^2
dy/dx=(2x^2+2-4x^2)/(x^2+1)^2
dy/dx=(2-2x^2)/(x^2+1)^2
Slope (m) of the tangent line at (1,1) = dy/dx at (1,1)
= ((2-2(1^2))/((1^2+1)^2))
= 0
Equation of the tangent line can be found by using point slope form
y-y_1 = m(x-x_1)
y-1= 0(x-1)
y-1=0
y=1
The normal line is perpendicular to the tangent line
Therefore slope of the normal line = -1/m
Equation of the normal is
y-y_1=(-1/m)(x-x_1)
m(y-y_1)=-(x-x_1)
0(y-1)=-(x-1)
x-1=0
x=1