Calculate the derivative indicated.

f"(0) , f(x) = 3x / e^x + `1

f '(x) = ((e^x+1)Dx(3x) - (3x)Dx(e^x +1)) / (e^x +1)^2

f '(x) = (3(e^x +1) - 3xe^x) / (e^x +1)^2

f '(x) = (3e^x + 3 - 3xe^x) / (e^x + 1)^2

since e^0 = 1; f '(0) =(3(1) +3 - 3(0)(1)) / (1 + 1)^2 = 6/4 = 3/2

By Quotient Rule :

[(e^x+1) * d/dx(3x) - 3x * d/dx (e^x+1) ] / (e^x+1)^2

[(3(e^x + 1)) - (3xe^x)] / (e^x+1)^2

f'(0) = [(3e^0+3)-(3(0)e^0)] /(e^0+1)^2

= 6/4