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Quickly Check my Work Please|Finding Derivatives

  1. Feb 15, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]y=\frac{8x^4-5x^2-2}{4x^3}[/tex]

    Find y'.

    2. The attempt at a solution
    [tex]y=\frac{8x^4-5x^2-2}{4x^3}[/tex]

    [tex]y'=\frac{(32x^3-10x)(4x^3)-(12x^2)(8x^4-5x^2-2)}{(4x^3)^2}[/tex]

    [tex]y'=\frac{128x^6-40x^4-96x^6+60x^4+24x^2}{16x^6}[/tex]

    [tex]y'=\frac{32x^6+20x^4+24x^2}{16x^6}[/tex]


    [tex]y'=\frac{4x^2(8x^4+5x^2+6)}{4x^2(4x^4)}[/tex]
    The 4x2 gets cancelled out.

    Finally,

    [tex]y'=\frac{8x^4+5x^2+6}{4x^4}[/tex]



    Did I do it right?
     
    Last edited: Feb 15, 2007
  2. jcsd
  3. Feb 15, 2007 #2
    Perrrfect.

    edit: The second and third terms in the numerator (of the final result) needs some corrections.
     
    Last edited: Feb 15, 2007
  4. Feb 15, 2007 #3
    Thank you very much!
     
  5. Feb 15, 2007 #4
    Rather than using the quotient rule, it would be quicker to simplify the original expression...

    [tex]y=\frac{8x^4-5x^2-2}{4x^3}[/tex]

    [tex]= 2x - \frac{5}{4} x^{-1} - \frac{1}{2}x^{-3}[/tex]

    Then, take the derivative. The form of the answer won't be the same as what you have, but it'll be equivalent.
     
    Last edited: Feb 15, 2007
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