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Homework Help: Find The Derivative

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    =[tex]\frac{-3x^{4}}{(4x-8)^{1/2}}[/tex]

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

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

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

    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:

    =[tex]\frac{-6x^{3}(8x-16-x)}{(4x-8)^{3/2}}[/tex]

    =[tex]\frac{-6x^{3}(7x-16)}{(4x-8)^{3/2}}[/tex]

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

    Answer:
    =[tex]\frac{-3x^{3}(7x-16)}{(4x-8)^{3/2}}[/tex]

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

    =[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)(4x-8)^{-1/2}(4)}{(4x-8)}[/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]

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

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

    =[tex]\frac{-6x^{3}(8x-16-x)}{(4x-8)^{3/2}}[/tex]

    =[tex]\frac{-6x^{3}(7x-16)}{(4x-8)^{3/2}}[/tex]

    The Answer:
    =[tex]\frac{-3x^{3}(7x-16)}{(4x-8)^{3/2}}[/tex]

    I hope this cleared it up a lot.
     
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