`"If "x=(2costheta-cos 2theta)and y=(2sin theta-sin 2theta)," find "((d^(2))/(dx^(2)))_(theta=(pi)/(2))`

If x=2cos theta -cos2 theta and y=2sin theta -sin2 theta , find (d^2y)/(dx^2) at theta=pi/2 .

If x = 2costheta- cos 2theta, y = 2sintheta -sin 2theta, find dy/dx.

If x =2 cos theta - cos 2 theta y = 2 sin theta - sin 2 theta Find (d^(2)y)/(dx^(2)) at theta = pi/2

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

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

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

if x =2 cos theta - cos 2 theta y = 2 sin theta - sin 2 theta Find (d^(2)y)/(dx^(2)) at theta = pi/2

Find dy/dx if x=costheta-cos2theta and y=sintheta-sin2theta

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

If x=2 cos theta-cos 2 theta and y=2 sin theta-sin 2 theta , then the value of (d^(2)y)/(dx^(2)) at theta=(pi)/(2) is -