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Mathematics 8 Online
OpenStudy (anonymous):

Differentiate implicitly, cos^2x+cos^2y=cos(2x+2y)

OpenStudy (anonymous):

\[-[\sin(2x)+\sin(2y)]\div 2=1/2 \sin(2x+2y)\]

OpenStudy (anonymous):

^thats rubbish lol

OpenStudy (anonymous):

anything wrong?

OpenStudy (anonymous):

yeh all of it

OpenStudy (anonymous):

you didnt find the derivative

OpenStudy (anonymous):

also the expression they have is \[\cos^2 x +\cos^2 y = \cos(2x+2y)\]

OpenStudy (anonymous):

you thought the ^2 , mean 2x

OpenStudy (anonymous):

differentiate both sides\[-\sin(2x) -\sin(2y) \frac{dy}{dx} = - ( 2 + 2\frac{dy}{dx}) \sin(2x+2y) \]

OpenStudy (anonymous):

also you integrated the terms :|, so many mistakes lol

OpenStudy (anonymous):

and the integral of cos is not - sin , lol what did u do

OpenStudy (anonymous):

rearrange my thing above to get dy/dx

OpenStudy (anonymous):

\[\frac{\text{dy}}{\text{dx}}=-\frac{\sin (x) \cos (x)-\sin (2 x+2 y)}{\sin (y) \cos (y)-\sin (2 x+2 y)} \]

OpenStudy (anonymous):

sorry you are right elecengineer

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