If x=cos(3t) and y=sin^2(3t), what is dy/dx and d^2y/dx^2

∵ x = cos 3t, y = sin² 3t

∴ x² + y = cos² 3t + sin² 3t = 1

∴ y = 1 - x² ........................................… (1)

∴ dy/dx = -2x ..................................... (2)

∴ d²y/dx² = -2 .................................... (3)
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Note that, with x = cos(3t) and y = sin^2(3t), we see that:
dx/dt = -3sin(3t) and dy/dt = 6sin(3t)cos(3t).

This gives:
dy/dx = (dy/dt)/(dx/dt)
= [6sin(3t)cos(3t)]/[-3sin(3t)]
= -2cos(3t).

Then, since d^2y/dx^2 = [(d/dt)(dy/dx)]/(dx/dt), we see that:
d^2y/dx^2 = [(d/dt)(dy/dx)]/(dx/dt)
= {(d/dt)[-2cos(3t)]}/[-3sin(3t)]
= [6sin(3t)]/[-3sin(3t)]
= -2.

I hope this helps!

dy/dx= -2cos(3t)

d^2y/dx^2= -2

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