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Simpe simplification.

  1. Dec 26, 2006 #1
    dy/dx, where [tex]y=\frac{e^{-2x}}{x^2}[/tex]


    Using the quotient rule I get

    [tex]\frac {-2e^{-2x}(x^2) - e^{-2x}(2x)}{x^4}[/tex]

    Simplified I get:-

    [tex]\frac{e^{-2x}(-2x^2-2x)}{x^4}[/tex]

    The answer is [tex]-\frac{2e^{-2x}(x+1)}{x^3}[/tex]

    Simple question can someone run me through the simplification, I'm not quite getting why it's over x^3 here. I'm sure it's just a simple fraction deal, but if someone could break it down nice and simply it would help.

    Thanks in advance.

    EDIT: sorry I corrected my second step. I accidently added an extra minus sign.
     
    Last edited: Dec 26, 2006
  2. jcsd
  3. Dec 26, 2006 #2
    You're pulling a 2x out from (2x2+ 2x).

    You need to keep track of the minus signs, though.
     
  4. Dec 26, 2006 #3
    Ah I see if I divide [tex]\frac{e^{-2x}(-2x^2-2x)}{x^4}[/tex] by -2x. I get [tex]-1/2x^3 (e^{-2x}(x+1))[/tex] which is [tex]-\frac {2e^{-2x}(x+1)}{x^3}[/tex] I'm wondering what my problem was here. Thanks there. Simple really.
     
    Last edited: Dec 26, 2006
  5. Dec 26, 2006 #4
    Actually, you divide AND multiply by -2x. :smile:
     
  6. Dec 26, 2006 #5
    Sure that's what I meant, thanks.:smile:
     
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