y= 1/ (x+2)(x-3)

logy= log1 - log(x+2) - log(x-3)

 1/y  (y')  = 0 - 1/(x+2) - 1/(x-3)

               = (-x+3-x-2)/(x+2)(x-3)

               = (-2x+1)/(x+2)(x-3)

 1/(x+2)(x-3)  y'    = (-2x+1)/(x+2)(x-3)

                       y'   = -2x+1 

Approved

dear vineeta

y= 1/ (x+2)(x-3)

or (x+2)(x-3) =1/y

    differtiate

  2x-1 =-1/y² dy/dx

or dy/dx  =-y2 (2x-1)

                    =-(2x-1)/(x+2)²(x-3)²