y = a^cos-1x/1+a^cos-1x. z=a^cos-1x,then dy/dx ?A.1/1+a^cos-1x. B.-1/a^cos-1x C.(1/1+ a^cos-1)^2

Dear Student,
the question must have asked dy/dz as z value is given.
y=a^cos-1x/(1+a^(cos^-1 x)) and z=a^(cos^-1 x)
 
dy/dx=( -a^(cos^-1 x)*log a*1*(1+a^(cos^-1 x))/(sqrt(1-x^2)) +
a^(cos^-1 x)*log a*a^(cos^-1 x)/(sqrt(1-x^2)))/(1+a^(cos^-1 x))^2
=((a^(cos^-1 x)*log a)*(-1-a^(cos^-1 x)+a^(cos^-1 x)))/((sqrt(1-x^2)*(1+a^(cos^-1 x)^2)
 
now, dz/dx =-a^(cos^-1 x)*log a /(sqrt(1-x^2))
so dy/dz= (dy/dx)/(dz/dx)= 1/(1+a^(cos^-1 x))^2
 
So C is the correct answer.