Question

Find y' y^2 = x cos y assuming that the equation determines a differentiable function f such that y= f(x)

2yy' = (1)(cosy) - (siny)y'(x)
2yy'-x(sin(y))y'=cosy
y'=cosy/(2y-xsiny)

Answer 1

just want to check my answer

Answer 2

2y dy/dx = x ( -sin(y) dy/dx ) +cos(y) )

Answer 3

wait, thats wrong

Answer 4

the cos9y) at the end shouldnt be in the bracket

Answer 5

cos(y)

Answer 6

kk its the same as my answer just in terms of dy/dx so its correct :)?

Answer 7

you did make a mistake

Answer 8

aww :(

Answer 9

an algebraic one, not a calculus one

Answer 10

hmm not suprised not a fan of algebra

Answer 11

first line to second line

Answer 12

you take a term from the RHS to the left, but dont change the sign

Answer 13

its suppose to be positive when i take it over?

Answer 14

yeah i just realized

Answer 15

ok thank you

Answer 16

ok thank you

Answer 17

also , just something notation wise,

the first line

2yy' = (1)(cosy) - (siny)y'(x) , is badly written

Answer 18

like I mean I could get what you mean, but be aware of the order of the terms

Answer 19

I know you mean to write -xsin(y) y' for the last term

Answer 20

but you have -sin(y) y'(x)

which can be interperated as -sin(y) dy/dx

Answer 21

it might confuse yourself and the markers, so I would be careful about that in the future , order your terms to eliminate confusion

Answer 22

oh i do it step by step i wrote it that way on my next line sorry

Answer 23

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

Answer 24

ok thanks for the advice i will remeber it for next time

Answer 25

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

Answer 26

ok