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Second derivative using implicit differentiation

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    x^6 + y^6 = -6

    I have to prove that y'' = 30x^4/y^11

    2. Relevant equations

    3. The attempt at a solution

    Using implicit differentiation:

    6x^5 + 6y^5 dy/dx = 0
    6y^5 dy/dx = -6x^5
    dy/dx = -x^5/y^5

    Quotient Rule:

    [(y^5)(-5x^4) - (-x^5)(5y^4 dy/dx)] / (y^5)^2
    [-5x^4 y^5 + (x^5) 5y^4 dy/dx] / (y^5)^2

    This is where I got stuck.
  2. jcsd
  3. Sep 29, 2011 #2


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    You can replace dy/dx again, and then simplify a bit.
    Also, you lost an equality sign somewhere... that expression on the last line, what is it equal to again?
  4. Sep 29, 2011 #3


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    Staff Emeritus
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    For me, rather than using the quotient rule, it was easier to do this:

    [tex] y^5 \frac{dy}{dx} = -x^5 [/tex]

    and differentiate this implicitly (using the product rule on the left-hand side). As CompuChip pointed out, you'll have to substitute in the expression for dy/dx, and you also get a handy substitution from the original statement that x^6 + y^6 = -6 .
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