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anonymous

  • 5 years ago

Find y' sinsqrt y -3x = 2 assuming that the equation determines a differentiable function f such that y=f(x)

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  1. anonymous
    • 5 years ago
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    you need to include brackets

  2. anonymous
    • 5 years ago
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    \[1/2 (\sin y)^{-1/2}(\cos y)y'-3=0\] that is what i got but i dont know how to find y' now i.e. simplify

  3. anonymous
    • 5 years ago
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    its not clear what the actual question is

  4. anonymous
    • 5 years ago
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    \[\sin \sqrt{y} - 3x=0\]

  5. anonymous
    • 5 years ago
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    yeh

  6. anonymous
    • 5 years ago
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    so sin( y^(1/2) ) -3x =2

  7. anonymous
    • 5 years ago
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    (1/2)y^(-1/2) dy/dx cos(y^(1/2) ) -3 =0

  8. anonymous
    • 5 years ago
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    or (sin y)^1/2

  9. anonymous
    • 5 years ago
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    \[\frac{1}{2\sqrt{y}} \frac{dy}{dx} \cos(\sqrt{y}) -3 =0 \]

  10. anonymous
    • 5 years ago
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    why'd we get rid of the sin? in the first term

  11. anonymous
    • 5 years ago
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    solve that

  12. anonymous
    • 5 years ago
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    why not ? lol

  13. anonymous
    • 5 years ago
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    derivative of sin(f (x) ) = f'(x) cos(f(x) ) yes?

  14. anonymous
    • 5 years ago
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    yes

  15. anonymous
    • 5 years ago
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    differentiate the inside, multiply and channge the sin to cos

  16. anonymous
    • 5 years ago
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    change

  17. anonymous
    • 5 years ago
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    ok thank you i understand that was my fall down

  18. anonymous
    • 5 years ago
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    you were thinking of \[\frac{1}{2} \sin^2 (\sqrt{y} ) -3x =2 \] I think

  19. anonymous
    • 5 years ago
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    or something similiar

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spraguer (Moderator)
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