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Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6

  1. Jun 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6


    2. Relevant equations
    n/a


    3. The attempt at a solution
    I'm using y'=(dy/dx)

    I found the derivative of the above problem.
    12x+8xy'+10y=0 (I used the product rule to find the derivative of 8xy')

    12x+8xy'+10y=0

    8xy'=-12x-10y

    xy'=(-12x-10y)/(8x)

    I checked the answer in the back of the book and it was y'=(-6x-4y)/(4x+y)

    I'm not sure where I missed a step or a calculation error

    Thanks!!
     
  2. jcsd
  3. Jun 27, 2008 #2
    You forgot the chain rule on your y^2 term.
     
  4. Jun 27, 2008 #3
    6x^2+8xy+y^2=6

    12x+8xy' +8y + 2y**=0

    This is wrong ...
     
  5. Jun 27, 2008 #4
    i see what you tried to do. u tried to combine the 8y and 2y(dy/dx). they cannot be added together.

    p.s.
    next time u get stuck on something like this u can always work backwards. you could take the book's answer and multiply the top and bottom by 2. then u can figure it out from there
     
  6. Jun 27, 2008 #5
    First of, thanks for all the help on my previous question, I have another one!

    1. The problem statement, all variables and given/known data
    Find dy/dx by implicit differentiation: 3x^2=(2-y)/(2+y)


    2. Relevant equations
    n/a


    3. The attempt at a solution
    I'm using y'=(dy/dx)

    I found the derivative of the above problem.
    First I used the quotient rule...
    6x=[(2+y)(y')-(2-y)(y')]/(2+y)^2

    6x=(2y'+yy'-2y'+yy')

    6x=2yy'/(2+yy')^2

    After this, I got into a jumbled mess, I wasn't sure if I had to add y' to the bottom of the quotient rule.


    I checked the answer in the back of the book and it was y'=-3x(2+y)^2/2


    Thanks!!
     
  7. Jun 27, 2008 #6

    dynamicsolo

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    Homework Helper

    Don't forget that the first term in the numerator involves the derivative of (2 - y), so the quotient should be

    [(2+y)(-y')-(2-y)(y')]/(2+y)^2 .

    You could multiply by (2+y)^2 to get

    6x · [(2+y)^2] = -2y' - yy' - 2y' + yy' = -4y'

    and solve for y' .
     
  8. Jun 27, 2008 #7

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Sign error: what's the derivitive of (2-y) ?

    What happened to the denominator on the RHS?

    How did (2+y)^2 from before become (2+yy')^2 ?
     
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