how to differentiate a logarithm

Question

ln[(1+sinx)/(1-sinx)]^(1/2)

Answer 1

y =ln ( ( 1+ sinx)/( 1- sinx))^1/2

                            1

dy/dx = -------------------------------- *d((1+sinx)/(1-sinx)^1/2

             ((1+ sinx)/( 1- sinx))^1/2

                                   d((1+sinx)/(1-sinx))

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

          ( sqrt((1+sinx)/(1-sinx)))*2(sqrt((1+sinx)/(1-sinx)))      

            [d(1+sinx)(1-sinx) -(1+sinx)*d(1-sinx)]/(1-sinx)^2

---------------------------------------------------------------------------  =

                        2(1+sinx)/(1-sinx)

         cosx(1-sinx) + cosx(1+sinx)

------------------------------------------------------------------ =

               2( 1+sinx)(1-sinx)

           cosx( 1-sinx+1+sinx)

-------------------------------------------- =

         2( 1-sinx^2)

cosx/(2socsx^2) = 1/2cosx