Derivatives of Trigonometric Functions: Solving for dy/dx and d²y/dx²

[JBD2]

Homework Statement

If x = cos(3t) and y = (sin(3t))² find dy/dx and d²y/dx².

Homework Equations

d/dx cosx = -sinx
d/dx sinx = cosx

The Attempt at a Solution

dx/dt = -3sin3t
dy/dt = 6sin3tcos3t

Not sure what to do now.

 

[Dick]

(dy/dt)/(dx/dt)=dy/dx, right?

 

[JBD2]

Yes, thank you. After I posted I wondered whether you could do that or not (a little unfamiliar with Leibniz notation) but it makes sense now thank you.