curve is given by the parametric equationx = 2 cos2(θ) and y = 3 sin(2θ).(i) Find y′(x) in terms of θ, simplifying your answer as far as possible

To find y′(x) in terms of θ, we first need to find dx/dθ and dy/dθ.

Given x = 2 cos²(θ), we differentiate with respect to θ to get dx/dθ. The derivative of cos²(θ) is -2sin(θ)cos(θ) using the chain rule. So, dx/dθ = -4sin(θ)cos(θ).

Similarly, given y = 3 sin(2θ), we differentiate with respect to θ to get dy/dθ. The derivative of sin(2θ) is 2cos(2θ) using the chain rule. So, dy/dθ = 6cos(2θ).

Now, we can find dy/dx = (dy/dθ) / (dx/dθ) = [6cos(2θ)] / [-4sin(θ)cos(θ)].

Simplify this to get dy/dx = -3/2 * [cos(2θ)] / [sin(θ)cos(θ)].

This is the derivative of y with respect to x, in terms of θ.