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Homework Help: Find the second derivative

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the second derivative of 3(x^2)y+y+x=x^5

    2. Relevant equations
    Find the first derivative using implicit differentiation.
    Find the second by using quotient rule.

    3. The attempt at a solution
    So I found the first derivative to be

    dy/dx = 5(x^4)-6xy-1 / 3(x^2)+1

    I get very lost after this trying to find the second derivative using the quotient rule as there is the y variable still in the equation. Any help please?
  2. jcsd
  3. Jul 21, 2012 #2


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    The equation for y' is wrong without parentheses.
    Isolate y from the first equation 3(x^2)y+y+x=x^5 and sub into y'.

  4. Jul 21, 2012 #3


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    Put u = 5(x^4)-6xy-1, then get du/dx using implicit differentiation and put v= 3(x^2)+1 and get dv/dx.

    Then just put it into your formula for the quotient rule. But in du/dx you will have a term with dy/dx in it (which you know from the first derivative).
  5. Jul 21, 2012 #4
    Finding du/dx I get 20(x^3)-6x(dy/dx)-6y and dv/dx I get 6x. Is that correct?

    This is what I tried earlier but got very lost trying to sub in dy/dx
    Last edited: Jul 21, 2012
  6. Jul 21, 2012 #5


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    It is correct now.:smile:

    Show your further work.

  7. Jul 21, 2012 #6
    so then using the quotient rule

    dy/dx = (v(du/dx)-u(dv/dx))/(v^2)

    = ((3(x^2)+1)*(20(x^3)-(6x((5(x^4)-6xy-1)) / (3(x^2)+1))-6y) - (5(x^4)-6xy-1)* (6x)) / ((3(x^2)+1)^2)

    sorry it looks very confusing like that if I knew how to make an image i would.

    My calculator gives a result of (2(10(x^3)+27(x^2)(y)+6x-3y) / (3(x^2)+1)^2)
    and I have not been able to get close to that answer.
  8. Jul 21, 2012 #7
    factoring the above I get a result of

    (-60(x^5)+20(x^3)+72(x^2)(y)+12x-6y) / ((3x^2+1)^2)


    (2(-30(x^5)+10(x^3)+36(x^2)(y)+6x-3y)) / ((3x^2+1)^2)
  9. Jul 21, 2012 #8


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    Finding y' by implicit diferentiation is a good idea and I see no reason not to do the same for the second derivative.

    [itex]3x^2y+y+x=x^5[/itex] so
    [itex]6xy+ 3x^2y'+ y'+ 1= 5x^4[/itex]

    Differentiating, implicitely, again,
    [itex]6y+ 6xy'+ 6xy'+ 3x^2y''+ y''= 20x^3[/itex]
    [itex]6y+ 12xy'+ (3x^2+ 1)y''= 20x^3[/itex]

    You can solve
    for [itex]y= \frac{x^5- x}{3x^2+ 1}[/itex] and
    [itex]6xy+ 3x^2y'+ y'+ 1= 5x^4[/itex]
    for [itex]y'= \frac{5x^4- 6xy+ 1}{3x^2+ 1}[/itex]
    and put those into [itex]y''= \frac{20x^3- 12xy'- 6y}{3x^2+ 1}[/itex]
    if you like but my experience is that you seldom have to solve for y'' in terms of x only.
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