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If #x = cos(3t)# and #y = sin^2(3t)#, how do you find dy/dx?

1 Answer

Apr 13, 2015

Since #sin^2x+cos^2x=1#

Than

#y=sin^2(3t)=1-cos^2(3t)=1-x^2#.

So:

#y'=-2x#.

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