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just want to check my answer
2y dy/dx = x ( -sin(y) dy/dx ) +cos(y) )
wait, thats wrong
the cos9y) at the end shouldnt be in the bracket
kk its the same as my answer just in terms of dy/dx so its correct :)?
you did make a mistake
an algebraic one, not a calculus one
hmm not suprised not a fan of algebra
first line to second line
you take a term from the RHS to the left, but dont change the sign
its suppose to be positive when i take it over?
yeah i just realized
ok thank you
ok thank you
also , just something notation wise, the first line 2yy' = (1)(cosy) - (siny)y'(x) , is badly written
like I mean I could get what you mean, but be aware of the order of the terms
I know you mean to write -xsin(y) y' for the last term
but you have -sin(y) y'(x) which can be interperated as -sin(y) dy/dx
it might confuse yourself and the markers, so I would be careful about that in the future , order your terms to eliminate confusion
oh i do it step by step i wrote it that way on my next line sorry
I dont care exactly what you do , I am just saying , I highly recommend you dont put y' and x next to eaach other in a term , especially with the x in a bracket, just remember that y'(x) is an alternative notation for dy/dx
ok thanks for the advice i will remeber it for next time
alternatively , you can just write dy/dx when you use chain rules instead of y' , I rarely write y' , because it doesnt tell what the derivative is with respect to , and once you start to do multivariable calculus then you actually do have to write dy/dx , because y' is meaningless when you have multiple variables