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Differentiate implicitly, cos^2x+cos^2y=cos(2x+2y)
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\[-[\sin(2x)+\sin(2y)]\div 2=1/2 \sin(2x+2y)\]
^thats rubbish lol
anything wrong?
yeh all of it
you didnt find the derivative
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also the expression they have is \[\cos^2 x +\cos^2 y = \cos(2x+2y)\]
you thought the ^2 , mean 2x
differentiate both sides\[-\sin(2x) -\sin(2y) \frac{dy}{dx} = - ( 2 + 2\frac{dy}{dx}) \sin(2x+2y) \]
also you integrated the terms :|, so many mistakes lol
and the integral of cos is not - sin , lol what did u do
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rearrange my thing above to get dy/dx
\[\frac{\text{dy}}{\text{dx}}=-\frac{\sin (x) \cos (x)-\sin (2 x+2 y)}{\sin (y) \cos (y)-\sin (2 x+2 y)} \]
sorry you are right elecengineer
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