`x=sqrt(sin 2theta), y=sqrt(cos2theta)` then find `(dy)/(dx)`

If x=a(2theta-sin2theta) and y=a(1-cos2theta) find (dy)/(dx) when theta = pi/3

If x=a(2 theta-sin2 theta) and y=a(1-cos2 theta) find (dy)/(dx) when theta=(pi)/(3)

If x=cos theta-cos 2theta and y=sin theta-sin2theta , then find (dy)/(dx) .

If y(x)=int_((pi^(2))/(16))^(x^(2))(cos x cos sqrt(theta))/(1+sin^(2)sqrt(theta)), then f in d(dy)/(dx)atx=pi

If x= a sin 2theta (1+ cos 2theta) , y= b cos 2theta(1- cos 2theta) , then (dy)/(dx) =

If x= a sin 2theta (1+ cos 2theta) , y= b cos 2theta(1- cos 2theta) , then (dy)/(dx) =

If x= a sin 2theta (1+ cos 2theta) , y= b cos 2theta(1- cos 2theta) , then (dy)/(dx) =

If x=sqrt(3)(3 sin theta+sin 3theta), y=sqrt(3)(3 sin theta+cos 3 theta) , find (d^(2)y)/(dx^(2)) at theta=(pi)/(3) .

x = a (cos theta + theta sin theta), y = a (sin theta - theta cos theta) . then find dy/dx