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#