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Calculus I - Double Derivitive

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx and d^2y/dx^2

    y = x / (x^(2)+1)


    2. Relevant equations

    d/dx (f/g) = (g d/dx f - f d/dx g) / g^2

    3. The attempt at a solution

    Finding d/dx:

    d/dx y = (x^(2)+1) d/dx (x) - (x) d/dx (x^(2) + 1) / (x^(2)+1)^2

    = (x^2 + 1) - (2x^2) / (x^(2)+1)^2

    So that's my first derivative answer.. now on to the second.

    d/dx(d/dx y) (x^2 + 1)^2 d/dx [(x^(2)+1)-(2x^2)] / (x^(2)+1)^4

    ((x^(2)+1)^2)(2x-4x)-[4x(x^(2)+1)-(2x^2)/ (x^(2)+1)^4


    So, there is bound to be a mistake somewhere.. thank you in advance
     
  2. jcsd
  3. Oct 27, 2009 #2

    lanedance

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    Homework Helper

    i find it quite difficult to read, as you leave out brackets,

    [tex] \frac{d}{dx} y = \frac{d}{dx} (\frac{x}{x^2 + 1}) [/tex]

    [tex] = \frac{(x^2 + 1)\frac{d}{dx} (x) - x \frac{d}{dx}(x^2 + 1)}{(x^2 + 1)^2} [/tex]

    [tex] = \frac{((x^2 + 1) - 2x^2) }{(x^2 + 1)^2} [/tex]
    looks similar to what you had, but with an extra bracket,

    though you should also simplify before differentiating again
    [tex] = \frac{ 1 - x^2 }{(x^2 + 1)^2} [/tex]



    as mentioned its hard to read without missing +,-,= & brackets

    that said I find wreting it as below, then diifferntiating again using the product rule a little easier, though it will lead to identical result as the quotient rule
    [tex] \frac{dy}{dx} = (1 - x^2)(x^2 + 1)^{-2} [/tex]
     
    Last edited: Oct 27, 2009
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