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anonymous

  • 5 years ago

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)

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  1. anonymous
    • 5 years ago
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    just want to check my answer

  2. anonymous
    • 5 years ago
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    2y dy/dx = x ( -sin(y) dy/dx ) +cos(y) )

  3. anonymous
    • 5 years ago
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    wait, thats wrong

  4. anonymous
    • 5 years ago
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    the cos9y) at the end shouldnt be in the bracket

  5. anonymous
    • 5 years ago
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    cos(y)

  6. anonymous
    • 5 years ago
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    kk its the same as my answer just in terms of dy/dx so its correct :)?

  7. anonymous
    • 5 years ago
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    you did make a mistake

  8. anonymous
    • 5 years ago
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    aww :(

  9. anonymous
    • 5 years ago
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    an algebraic one, not a calculus one

  10. anonymous
    • 5 years ago
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    hmm not suprised not a fan of algebra

  11. anonymous
    • 5 years ago
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    first line to second line

  12. anonymous
    • 5 years ago
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    you take a term from the RHS to the left, but dont change the sign

  13. anonymous
    • 5 years ago
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    its suppose to be positive when i take it over?

  14. anonymous
    • 5 years ago
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    yeah i just realized

  15. anonymous
    • 5 years ago
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    ok thank you

  16. anonymous
    • 5 years ago
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    ok thank you

  17. anonymous
    • 5 years ago
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    also , just something notation wise, the first line 2yy' = (1)(cosy) - (siny)y'(x) , is badly written

  18. anonymous
    • 5 years ago
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    like I mean I could get what you mean, but be aware of the order of the terms

  19. anonymous
    • 5 years ago
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    I know you mean to write -xsin(y) y' for the last term

  20. anonymous
    • 5 years ago
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    but you have -sin(y) y'(x) which can be interperated as -sin(y) dy/dx

  21. anonymous
    • 5 years ago
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    it might confuse yourself and the markers, so I would be careful about that in the future , order your terms to eliminate confusion

  22. anonymous
    • 5 years ago
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    oh i do it step by step i wrote it that way on my next line sorry

  23. anonymous
    • 5 years ago
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    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

  24. anonymous
    • 5 years ago
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    ok thanks for the advice i will remeber it for next time

  25. anonymous
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    ok

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