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Implici differentiation

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    dy/dx: square root x+y= 1+x^3y^2

    2. Relevant equations

    chain rule
    implicit differentiation

    3. The attempt at a solution

    1/2 x+y -1/2 =2x^2y^3 *y'
     
    Last edited: Jul 6, 2008
  2. jcsd
  3. Jul 6, 2008 #2
    [tex]\sqrt{x+y}=1+x^3y^2[/tex]

    Be clear!!! Make use of parenthesis. Right?
     
  4. Jul 6, 2008 #3
    yeah, my bad
     
  5. Jul 6, 2008 #4
    1/2(x+y)^-1/2(x+y)'= (2x^2y^2)(y)'(x^3)
     
  6. Jul 6, 2008 #5
    am i going in the right direction?
     
  7. Jul 6, 2008 #6

    rock.freak667

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    To differentiate the LHS w.r.t x
    1/2(x+y)^-1/2 is correct but you'll need to multiply it by the differential of (x+y) i.e. what is in the bracket.

    For the RHS : [itex]1+x^3y^2[/itex] use the product law for [itex]x^3y^2[/itex]
     
  8. Jul 6, 2008 #7
    ok
    1/2(x+y)^-1/2 + 1/2(x+y)^-1/2 (y)'= 3x^2y^2 +2y (y)' (x^3)
     
  9. Jul 6, 2008 #8
    is that right?
     
  10. Jul 6, 2008 #9

    Defennder

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    Yes it is.
     
  11. Jul 6, 2008 #10
    do i need to simplify anymore?
     
  12. Jul 6, 2008 #11

    Defennder

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    Are you required to?
     
  13. Jul 6, 2008 #12
    I need to find y'
     
  14. Jul 6, 2008 #13
    is this the answer?
    y'= 3x^2-1/2(x+y)^-1/2 over [1/2(x+y)^-1/2] - 2yx^3
     
  15. Jul 6, 2008 #14
    Defennder confirmed your "Calculus steps" I'm sure you can handle the rest.
     
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