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How to use the general for of the derivative formula with a fractional exponent

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    f(x)=x^7/2, Find the derivative


    2. Relevant equations

    f'(a)=limit as x approaches a; ((x+h)-x)/h

    3. The attempt at a solution
    limit as h approached 0; ((x+h)^7/2-x^7/2)/h
    Multiplied by the conjugate with -5/2 as exponent and came up with 2x^5/2, which is incorrect
     
  2. jcsd
  3. Sep 23, 2012 #2
    do you want to evaluate [itex] \frac{x^7}{2}[/itex] or[itex] (x/2)^7 [/itex]?
    in the first case, we get [itex] (\frac{(x+h)^7}{2} - \frac{x^7}{2})/h [/itex]. evaluating (x+h)^7 would require the binomial theorem, which says that [itex](x+y)^n = Ʃfrom k to n \frac{n!}{k!(n-k)!}x^{n-k}.y^k [/itex]
     
  4. Sep 23, 2012 #3

    Dick

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    The conjugate of (x+h)^(7/2)-x^(7/2) is (x+h)^(7/2)+x^(7/2). Multiplying numerator and denominator by that is a good first step. Can you please show what you did?
     
  5. Sep 23, 2012 #4
    Okay we were multiplying it by (x+h)^(-5/2)+x^(-5/2), let me try it again
     
    Last edited: Sep 23, 2012
  6. Sep 23, 2012 #5
    Hmmm, okay this is a mess, how do deal with this mess (x+h)^7-x^7?
     
  7. Sep 23, 2012 #6

    Dick

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    Use the binomial theorem or just multiply it out. You will only wind up caring about the terms with highest power in x. Can you see why?
     
  8. Sep 23, 2012 #7
    Im looking at it but dont see it
     
  9. Sep 23, 2012 #8

    Dick

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    (x+h)^7=x^7+7*x^6*h+21*x^5*h^2+.... I'm using the binomial theorem. Why don't you care about terms that have h^2 or higher powers of h in them? Remember there's an h in the denominator and h will eventually go to zero.
     
  10. Sep 23, 2012 #9
    I get that h goes to zero, but I thought we had to cancel it out first? Can we just assume zero for each term with h in it?
     
  11. Sep 23, 2012 #10

    Dick

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    No. Look, you really have to write out what you have gotten so far. Show your work.
     
  12. Sep 23, 2012 #11
    Okay,

    [(x+h)^7-x^7]/h((x+h)^(7/2)+x^7)=

    [x^7+7x^6h+21x^5h^2+35x^4h^3+35x^3h^4+21x^2h^5+7xh^6+h^7-x^7]/h[(x+h)^(7/2)+x^(7/2)]
     
  13. Sep 23, 2012 #12

    Dick

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    Right. Thanks. You've got it. Cancel the x^7 in the numerator. Then divide the h in the denominator into the numerator. Now let h->0.
     
  14. Sep 23, 2012 #13
    Holy cow, that was a mess, ha ha!
    It works out now. I was helping a friend with her calc and haven't done this in a little while. Looks like I need to review!

    Thanks Dick!
     
  15. Sep 23, 2012 #14

    Dick

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    Great! You see now that you only need to write (x+h)^7=x^7+7*x^6*h+... You don't need to care about the terms represented by the '...'. Right?
     
  16. Sep 23, 2012 #15
    Yup, got it, just like the others. First time doing the general form with fractional exponents, threw me through a loop, must be spending too much time in physics lol
     
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