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Homework Help: Calculating derivatives (2 questions)

  1. Sep 12, 2009 #1
    Hello! I took an exam awhile back and missed two questions but have no feedback describing what went wrong. Let me know what you think.

    1. The problem statement, all variables and given/known data

    Derivative of x / 1 - x

    Second derivative of -1/3x

    2. The attempt at a solution

    My answer for 1 was the number -

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

    1-x / (1-x)^2 - Cancel leaving x/1-x

    x / -x = -1

    -1 / 1 = -1


    Second problem:

    f''(x) -1/3x
    -1/3x = -3x^-1
    f'(x) = (-1)(-3x)^-2 = 1/3x^2
    f''(x) = (-2)(3x)^-3 = 1/-6x^3

    These two problems are incorrect, I'm not sure why.
  2. jcsd
  3. Sep 12, 2009 #2
    Hi There

    You have a sign error I noticed for #1 - the derivative is indeed (1)(1-x) - x(-1) but that is not 1-x it is 2-x

    So the derivate would be (2-x)/(1-x)^2

    For your second problem, your first derivative is right = 1/3x^2
    now for the second, use the quotient rule again.

    f''x = [0 - (1)(6x)]/[(3x^2)^2)]
    the denominator there would be 9x^4, so the 2nd derivative simplifies to -2x/3x^4
  4. Sep 12, 2009 #3
    If you have [tex]f(x) = \frac{x}{1 - x}[/tex] then:

    f'(x) = \frac{(1)(1-x) - x(-1)}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}

    For the second problem, which I believe is [tex]g(x) = -\frac{1}{3x} = -\frac{1}{3}\ast\frac{1}{x} = -\frac{1}{3}x^{-1}[/tex], then you just use the chain rule to get:

    g'(x) = -\frac{1}{3}(-1)x^{-2} = \frac{1}{3}x^{-2}
    g''(x) = \frac{1}{3}(-2)x^{-3} = -\frac{2}{3}x^{-3}
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