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Derivative Problem

  1. Dec 5, 2008 #1
    2. Relevant equations
    find
    f'(x)= [(x^2)(x-3)]^(1/3)



    3. The attempt at a solution

    i got (3(x-2))/[x(x-3)^(2/3)]

    is that correct? im so confused if i got it right or not.

    Thanks before hand
     
  2. jcsd
  3. Dec 5, 2008 #2
    Not quite. I'd start by making it simpler by multiplying out the polynomials. Then you'd have:

    [tex](x^{3}-3x^{2})^{\frac{1}{3}}[/tex]

    Then, use the chain rule:

    If f(x)=g(h(x)), then f ' (x) = g ' (f(x))*f ' (x).

    For example, look at f(x) = (sin(x))2.

    So if g(x) = x2, and h(x) = sin(x), then f(x)=g(h(x)). Then to take the derivative, first, find the derivative of g and h.

    So g(x)=x2, which means that g ' (x)=2x.
    h(x) = sin(x), which means that h ' (x)= cos(x).

    Then to take the derivative, you would remember that f ' (x) = g ' (f(x))*f ' (x). So first, look at g ' (x). That's 2x. So put f(x) in for x, to get 2(sin(x)). Then, multiply by f ' (x), which is cos(x).

    So f ' (x), in this example, is 2(sin(x))*cos(x).

    Now, use that same method for your problem.
     
  4. Dec 5, 2008 #3
    This is an application of the chain rule.. You can multiply the inner product and get...[x^3-3x^2]

    ... now you bring the 1/3 out in front and reduce it by 1 as by application of the derivative. 1/3-1/1 = -2/3

    So you get what 1/3[d/dx[x^3-3x^2]]^-2/3

    so then it's: 1/3[x^2*(x-3)]^(-2/3) * derivative of the inside [3x^2-6x]

    Got it?
     
  5. Dec 5, 2008 #4

    Mentallic

    User Avatar
    Homework Helper

    never mind -.-
     
  6. Dec 5, 2008 #5

    Mark44

    Staff: Mentor

    If you're trying to find f'(x), look no further than the equation you wrote.

    I'm being facetious here. What you most likely meant was:
    f(x)= [(x^2)(x-3)]^(1/3)
    Find f'(x).

    This might seem like a small difference, but it's important to keep a function and its derivatives clearly separated in your mind.

     
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