Find the derivative f(X)=ln(2x/x+3)

Question

I would like to know how to get the answer too if possible, thanks

Answer 1

f(x) = ln(2x/x + 3)

                       1

dy/dx = ------------------------- * d(2x/(x+3))

                  2x/(x+3)

                      1               d(2x)*(x + 3) - 2x * d(x+3)

dy/dx = ---------------- * -------------------------------------      [ d(x+3) =1]

                2x/(x+3)                       ( x + 3 )^2

                    2(x+3) - 2x

dy/dx = ----------------------------------------------

                            2x(x+3)

                     2x + 6 - 2x

dy/dx =   -----------------------------

                      2x(x+3)

dy/dx = 6/(2x(x+3))

Answer 2

given f(x)=ln 2x/(x+3)

by appliying product rule 

let,2x/(x+3)=u

so

    y=lnu

   dy/du=1/u

           =1/(2x)/(x+3)

           =(x+3)/2x

du/dx={(x+3)2-2x}/(x+3)^2

        =6/(x+3)^2

but

 dy/dx = dy/du *du/dx

         (x+3)/2*6/(x+3)^2

      =6/2x(x+3)