Consider x = 0 , y = 1
Using Quotient Rule of Differentiation:
Let u = 2e^x and v = 1 + e^x
dy/dx = (vu' - uv')/v^2 , and given that d e^x/dx = e^x
dy/dx = ((1+e^x)(2e^(x)) - (2e^x)(e^(x)))/ (1+e^x)^2
dy/dx = (cancel((1+e^x))(2e^(x) ) - (2e^x)(e^(x)))/ ((1+e^x)^cancel(2))
dy/dx = ((2e^(x) )/ (1+e^x)) - ((2e^x)(e^(x)))/ (1+e^x)^2)
When x = 0
dy/dx = [(2e^(0))/(1+e^0)] - {([(2e^0)] [e^(0)])/((1+e^0)^2)}
dy/dx = (2(1))/((1+1)) - ((2(1)(1))/((1+1)^2))
dy/dx = 2/2 -2/2^2
dy/dx = 1 - 1/2
dy/dx = 1/2
So the slope at x = 0 is 1/2 (f'(x) = m = 1/2).
Consider:
y - y_0 = m(x-x_0)
y_0 = 1, x_0 = 0 ( this is at the point (0,1) )
y - f(x_0) = f'(x_0)(x-x_0)
y-1=1/2(x-0)
y = 1/2x +1
So the equation of the tangent line at (0,1) is y=1/2x +1