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Explanation of a simple exponent rule in a derivative problem needed

  1. Aug 20, 2011 #1
    Explanation of a "simple" exponent rule in a derivative problem needed

    After differentiation, I get this : 9(t-2/2t+1)8 * 5/(2t+1)2

    Now this simplifies into 45(t-2)/(2t+1)10

    Now, I am wondering what the property is that combines the exponent from the top with the bottom to get an power of 10 in the simplified form.

    *Note. the t-2/2t+1 is t-2 over 2t+1, the 5/(2t+1)2 is 5 over (2t+1)2, and the 45(t-2)/(2t+1)10 is 45(t-2) over (2t+1)10.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 20, 2011 #2
    Re: Explanation of a "simple" exponent rule in a derivative problem needed

    From the rules of exponents:

    [itex]a^n a^m = a^{n+m}[/itex]
     
  4. Aug 20, 2011 #3
    Re: Explanation of a "simple" exponent rule in a derivative problem needed

    if you start with [tex] 9 (\frac{t-2}{2t+1})^8 * \frac{5}{(2t+1)^2} [/tex]
    then you can write [tex] 9 \frac{(t-2)^8}{(2t+1)^8}* \frac{5}{(2t+1)^2} [/tex]
    and then you add the exponents when multiplying like bases, thus getting [tex] 45 \frac{(t-2)^8}{(2t+1)^{10}} [/tex]

    Note that [tex] (2t+1)^8(2t+1)^2=(2t+1)^{10} [/tex]

    edit:
    boo, pibond beat me cause I'm slow at latex! :P
     
    Last edited: Aug 20, 2011
  5. Sep 11, 2011 #4
    Re: Explanation of a "simple" exponent rule in a derivative problem needed

    Thank you. I was aware of the rule for exponents, but applying the power of 8 to both the numerator and denominator escaped me, now it all makes sense!
     
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