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Derivative of f(x)=(square root x^2-2x)^3-9(square rootx^2-2x)

  1. Jun 7, 2005 #1
    How do I find the derivative of f(x)=(square root x^2-2x)^3-9(square rootx^2-2x)
     
  2. jcsd
  3. Jun 7, 2005 #2

    mathwonk

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    use the prower rule and the chain rule. they are in your book or online.
     
  4. Jun 8, 2005 #3
    [tex]f(x) = (\sqrt{x^2 - 2x})^3 - 9\sqrt{x^2 - 2x}[/tex]

    Just like what Mathwonk said, use the chain rule for the first one
    [tex] (\sqrt{x^2 - 2x})^3[/tex]

    [tex] \sqrt{x^2 -2x} [/tex] is the inside function g(x) and [tex] x^3 [/tex] is the outer function f(x)

    The chain rule states, [tex] h '(x) = f '(g(x)) * g'(x) [/tex]



    For the power rule, for
    [tex]f(x)g(x) [/tex], the derivative is [tex]f'(x)g(x) + g'(x)f(x)[/tex]

    I'm giving you this information because you can find it in your book. Its up to you to find those inside and outer functions and differentiating them
     
    Last edited: Jun 8, 2005
  5. Jun 8, 2005 #4
    That's the product rule. Power rule:

    [tex]\frac{d}{dx} x^n = nx^{n - 1}[/tex]
     
  6. Jun 8, 2005 #5

    HallsofIvy

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    Of course, to use the power rule you will need to know that [tex]\sqrt{x}= x^{\frac{1}{2}}[/tex].
     
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