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Find The Derivative

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Is it actually correct, I'm not sure if it's correct, still.

    2. Relevant equations

    Quotient Rule and Chain Rule

    3. The attempt at a solution




    I'm not sure if it's correct up to here, but the [tex](4x-8)^{1/2}[/tex] isn't working. If there was no [tex]^{1/2}[/tex] it would work something like this:



    But still unsure how they get the -3 in front and not -6 like I got.


    Any help is appreciated. Thank you!

    Look 2 posts lower for CLEARED UP version!

    Still looking for help!
    Last edited: Feb 27, 2010
  2. jcsd
  3. Feb 27, 2010 #2
    In '3. The attempt at a solution' - I am not sure how you went from the second last to last equation (before you say "I'm not sure if it's correct up to here"). I have not gone through the whole thing.
  4. Feb 27, 2010 #3
    To the OP, I think your answer is correct, whereas whatever the answer at the very end is in fact incorrect. Another way to see this is to multiply the top and bottom of
    by (4x-8)^(1/2). The numerator of the resulting expression simplifies to -12x^3(4x-8) + 6x^4 which after more algebra comes down to your expression.
  5. Feb 27, 2010 #4
    Ok, starting from 3 again:

    All I'm using is the Quotient Rule here.


    =[tex]\frac{(-12x^{3})(4x-8)^{1/2} - (-3x^{4})(1/2)(4x-8)^{-1/2}(4)}{[(4x-8)^{1/2}]^{2}}[/tex]


    =[tex]\frac{-12x^{3}(4x-8)^{1/2} + 3x^{4}(1/2)(4)(4x-8)^{-1/2}}{(4x-8)}[/tex]

    =[tex]\frac{-12x^{3}(4x-8)^{1/2} + 3x^{4}(2)(4x-8)^{-1/2}}{(4x-8)}[/tex]

    =[tex]\frac{-12x^{3}(4x-8)^{1/2} + 6x^{4}(4x-8)^{-1/2}}{(4x-8)}[/tex]

    Right here, I bring down the [tex](4x-8)^{-1/2}[/tex], to make it positive [tex](4x-8)^{1/2}[/tex]

    =[tex]\frac{-12x^{3}(4x-8)^{1/2} + 6x^{4}}{(4x-8)(4x-8)^{1/2}}[/tex]

    =[tex]\frac{-12x^{3}(4x-8)^{1/2} + 6x^{4}}{(4x-8)^{3/2}}[/tex]


    Then the rest... (which is apparently wrong somewhere in the question)



    The Answer:

    I hope this cleared it up a lot.
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