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

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Answer 1 · 2015-04-13T17:22:42

Since ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

Than

$y = {sin}^{2} (3 t) = 1 - {cos}^{2} (3 t) = 1 - x^{2}$ $y=sin^2(3t)=1-cos^2(3t)=1-x^2$ .

So:

$y'=-2x$ .

If x = cos(3t)x=cos(3t)x = cos(3t) and y = sin^2(3t)y=sin2(3t)y = sin^2(3t), how do you find dy/dx?