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Calculus: Second derivative

  1. Feb 26, 2005 #1
    Find the second derivative by implicit differentiation:

    [tex] x^3 + y^3 = 6xy [/tex]

    Wow, a lengthy question when you have to show all your work. Anyway, for the first derivative I got:

    [tex] \frac {2y-x^2}{y^2-2x} [/tex]

    How would one start to calculate the second derivative? I tried the Quotient Rule and got:

    [tex] \frac {(y^2-2x)(2\frac{dy}{dx}-2x) - (2y-x^2)(2y\frac{dy}{dx}-2x)}{(y^2-2x)^2} [/tex]

    I'm not too sure what to do now.Would it be easier to bring the denominator to the top to get [tex] (2y-x^2)(y^2-2x)^-1 [/tex] and then using the chain and product rule? Thanks in advance.
  2. jcsd
  3. Feb 26, 2005 #2


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    Homework Helper

    Don't make the first derivative look like a quotient yet. Keep it in the form of :

    [tex]\frac{dy}{dx}(y^2 - 2x) = 2y - x^2[/tex]

    and differentiate both sides implicitly. Use the product rule and the chain rule.

    You'll get the second derivative as an expression in terms of the first derivative, x and y. You need to substitute in the expression you found for the first derivative (the quotient) at this point and simplify. Keep that simplification to the last step. Your final answer should be in terms of x and y only.

    EDIT : Although, honestly, even doing it the way you did it is fine. Just take that expression (after the quotient rule) and substitute in the expression you already have for the first derivative and simplify till you get an answer in x and y.
    Last edited: Feb 26, 2005
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