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Using The Power Rule

  1. Jun 10, 2006 #1
    I tried this derivative problem, but the back of the book shows a different answer then what I got. Can someone explain what I'm doing wrong.

    [tex]3T^5 -5T^.5 + \frac{7}{T}[/tex]

    So I did this:

    [tex]15T^4 - 2.5T^-.5 - 7T^-8 [/tex]

    It's the last part I got wrong I'm not sure why. I converted 7/T to T^-7. Is that right?
     
  2. jcsd
  3. Jun 10, 2006 #2

    Hootenanny

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    Not quite;

    [tex]\frac{7}{T} = 7T^{-1}[/tex]
     
  4. Jun 10, 2006 #3
    hmm, it's not doing the exponents right, u might have too look at the code.
     
  5. Jun 10, 2006 #4
    Oh, does the -1 come from the 7 or T?
     
  6. Jun 10, 2006 #5

    Hootenanny

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    It comes from the T, remember;

    [tex]\frac{1}{T} = T^{-1}[/tex]
     
  7. Jun 10, 2006 #6
    So [tex] \frac {1}{2T} = 1T^{-2}[/tex]?
     
  8. Jun 10, 2006 #7

    dav2008

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    See below message.
     
  9. Jun 10, 2006 #8

    Hootenanny

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    Nope, not quite.

    [tex]\frac{1}{2T} = (2T)^{-1}[/tex]

    The power applies to all terms of the denominator. However;

    [tex]\frac{1}{T^2} = T^{-2}[/tex]

    Does that make sense?
     
  10. Jun 10, 2006 #9
    So, you basically just combine the top and bottom terms and negate the exponent.
     
  11. Jun 10, 2006 #10

    Hootenanny

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    The exponent is the most important part. You can't just negate it.
     
  12. Jun 10, 2006 #11
    Ok, well I had this other problem Maybe I can get it right here.

    [tex]\frac{t^2 + t^3 -{1}}{t^4}[/tex]

    I divided by T^4 and then used the power rule and I got: [tex] .5T^{-.5} + .75T^{-.25} + 4T^{-5}[/tex]
     
    Last edited: Jun 10, 2006
  13. Jun 10, 2006 #12

    dav2008

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  14. Jun 10, 2006 #13
    Thanks, I got it.
     
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