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differentiate
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sin(x+y)=sin(xy) implicit
Cos(x+y)(1+y')=Cos(xy)(y+y'x)
Solve for y'
sorry sin(x+y)=1+sin(xy)
Same answer
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derivative of 1 is 0 lol
shift all y to LHS y' = [sin(y)-sin(x+y)]/[sin(x+y)+xsin(y)] xy'sin(y) ans.
cos(x+y)(1+y')=0 + cos(xy)(xy')(y)
cos(x+y)(1+y')=ycos(xy)(xy')
no the left side is Cos(xy)(y+y'x)
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use product rule
on the left?
yea for (x*y)' = x*y' + y*1
oh thats the right?
Oh yea i meant right sorry
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cos(x+y)(1+y')=cos(xy)(xy'+y)
now I have to get y' alone
hate this part :(
lol same here that's y i didn't do it hahah
hello :)
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cos(x+y)(1+y')=cos(xy)(xy'+y) now im trying to get y' alone
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