#y=sqrt(1-9x^2)arccos(3x)#
Let #" "3x=costheta#
So Differentiating w r to x we have #" " 3=-sintheta(d theta)/(dx)#
#=>(d theta)/(dx)=-3/sintheta#
Now
#y=sqrt(1-9x^2)arccos(3x)#
#=>y=sqrt(1-cos^2theta)arccos(costheta)#
#=>y=thetasintheta#
Differentiating w r to #theta# we have
#(dy)/(d theta)=thetacostheta+sintheta#
So
#(dy)/(dx)=(dy)/(d theta)xx(d theta)/(dx)#
#=>(dy)/(dx)=(thetacostheta+sintheta)xx(-3/sintheta)#
#=>(dy)/(dx)=(-3thetacostheta/sintheta-3)#
#=>(dy)/(dx)=-3((thetacostheta)/sqrt(1-cos^2theta)+1)#
#=>(dy)/(dx)=-3((3xcos^-1(3x))/sqrt(1-9x^2)+1)#
