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Derivative of y = x (1 - x^2)^1/2, is this correct?

  1. Nov 14, 2004 #1
    I keep on getting a weird answer when i take the dy/dx for this...

    i got x(1/x^2)^(-1/2)*-2x + (1-x^2)^(1/2)

    ... did i do that correctly?
    Last edited: Nov 14, 2004
  2. jcsd
  3. Nov 14, 2004 #2


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

    Use product rule/chain rule:

    y=(x)(\frac{1}{2})(\frac{1}{\sqrt{1-x^2}})(-2x) + (1-x^2)^\frac{1}{2}

    Slightly more simplified:

    y=(\frac{-x^2}{\sqrt{1-x^2}}) + (1-x^2)^\frac{1}{2}

    Pretty sure thats the answer...if so you aren't using the chain rule correctly. Double check how you get the derivitive of [tex] (1-x^2)^\frac{1}{2} [/tex]
  4. Nov 14, 2004 #3
    DUH. *slaps self on forhead* .... heh... forgot a single step and it screwed me up (of course).

    Last edited: Nov 14, 2004
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