$\displaystyle cos^2x+cos^2y = cos(2x+2y) $ $\displaystyle -2cos(x)sin(x)-2sin(y)y'cos(y) = -sin(2x+2y)(2+2y') $ //so far so good?
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Originally Posted by rabert1 $\displaystyle cos^2x+cos^2y = cos(2x+2y) $ $\displaystyle -2cos(x)sin(x)-2sin(y)y'cos(y) = -sin(2x+2y)(2+2y') $ //so far so good? It looks fine so far, so now solve for $\displaystyle \displaystyle \begin{align*} y' \end{align*}$
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