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Homework Help: Differentiate the function (derivatives, difference of sums rule)

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex] f(x) = x^{1/2} - x^{1/3} [/itex]

    2. Relevant equations
    [itex] f(x) = f'(x)- g'(x)[/itex]

    3. The attempt at a solution
    I am a little stuck about what to do after the first couple steps. Here is my attempt.

    [itex] f(x) = x^{1/2} - x^{1/3}[/itex]
    [itex]f'(x) = (x^{1/2})' -(x^{1/3})'[/itex]
    [itex]= 1/2x^{-1/2} - 1/3x^{-2/3}[/itex]
    **** this is where I get confused *****
    [itex]= 1/2x^{1/2} - 1/3x^{2/3}[/itex]
    [itex]= ?[/itex]

    Ps, I was recently told by a mentor not to write the prime in my solution, however, I'm leaving it like this because this is how it was taught in the course and I'm trying to avoid confusing myself further. Thank you.
  2. jcsd
  3. May 7, 2013 #2
    wat is confusing about it? you got the right answer halfway through. how would you write x^-1/2? remember that x1/2 is of course [itex]\sqrt[]{x}[/itex]. so x-1/2 would be 1 over that: 1/[itex]\sqrt[]{x}[/itex]. same concept remember with x-1 = 1/x.

    maybe the power of x is confusing you. remember that x3/4 for instance is the same as [itex]\sqrt[4]{x^3}[/itex].

    so you could also write it as 0.5/[itex]\sqrt[]{x}[/itex] - 0.333/[itex]\sqrt[3]{x^2}[/itex]
    Last edited: May 7, 2013
  4. May 7, 2013 #3
    Oh okay I think I understand. Thank you :)
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