find dy/dx and (d^2)y/(d^2)x for x=2sint y=3cost 0<1<2pi

Answer 1

dy/dx = y'/x' = (-3sint)/(2cost) = -3/2 tan t

d^2y/dx^2 = (x'y"-x"y')/x'^3
= ((2cost)(-3cost) - (-2sint)(-3sint))/(8cos^3 t)
= -3/4 sec^3 t

Check:
y = 3/2 √(4-x^2)
y' = -3/2 x/√(4-x^2)
y" = -6/(4-x^2)^3/2

I'll leave it to you to verify that they are the same.