Y =   ,     

dy/dx can be either +1 or -1.

we know cos-1 (cosx)=x ,(pi+x),-x,(-pi-x) depending on the value of x.here we dont know anything about x.(more precisely alpha-x)

using basic trigo)we get y=cos-1(cos *(alpha-x)  or cos-1cos(x-alpha)=alpha -x ,x-alpha etc .etc as above.

here alpha= tan-1(3/2)

on diffrentating we get dy/dx=+1 or -1 (.: alpha is a constant) 

 y = cos^-1 [ (2cosx  +  3sinx ) / ] 

now let  2 /  sqrt(13) = cos

therefore, 3 / sqrt(13) =sin

thus y = cos^-1 [ cosx cos  + sinx sin ]

y = cos^-1 cos ( x -  )------------1

y = x -  

now differentiating.........

y = 1.......but this cud be -1  if we write eq 1 as y = cos^-1 cos (  - x )

hope u got it.............

rate is useful.........