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Derivative of 1/(4x^2 + 3x - x)

  1. Apr 30, 2004 #1
    what is the derivative of 1/(4x^2+3x-x)

    I got 1/4x is that correct?
     
  2. jcsd
  3. Apr 30, 2004 #2

    Doc Al

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    Staff: Mentor

    How did you get that?
     
  4. Apr 30, 2004 #3
    1/(4x^2+3x-x) =
    (1/4x^2)+(1/3x)-(1/x)=
    ((4x^2)^-1 )+ ((3x)^-1) - ((x)^-1))=
    4x+3-1/x
    derivative = 4x^-1 = 1/4x
     
  5. Apr 30, 2004 #4
    Wrong.

    The answer is -(4x+1)/((2x^2)*(2x+1)^2)
     
  6. Apr 30, 2004 #5

    Doc Al

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    Staff: Mentor

    1/(a+b) does not equal 1/a + 1/b !!

    Yikes! :eek:
    If that were true, then this would follow:
    [tex]\frac{1}{2} = \frac{1}{1+1} = \frac{1}{1} + \frac{1}{1} = 2[/tex]
     
  7. Apr 30, 2004 #6
    oh wow, what was I thinking??!?!? whooops. How would I solve this then?
     
  8. Apr 30, 2004 #7
    You could use the quotient rule, or simply rewrite the equation as [itex](4x^2+2x)^{-1}[/itex] and use the regular chain rule.

    PS: [itex]\frac{1}{A+B+C}[/itex] does not equal [itex]\frac{1}{A}+\frac{1}{B}+\frac{1}{C}[/itex] ; however, [itex]\frac{A+B+C}{D}[/itex] does equal [itex]\frac{A}{D}+\frac{B}{D}+\frac{C}{D}[/itex]. See the difference? You cant simply split a denominator but you can split up a numerator. Use partial fractions to split the denominator.

    [edit] fixed my tex tags
     
    Last edited: Apr 30, 2004
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