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Given f(x) = x^(2/3), find f'(x) using the Δ method.

  1. Dec 17, 2013 #1


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    2. Relevant equations
    f'(x) = (f(x + Δx) – f(x))/(Δx)

    3. The attempt at a solution
    I understand the entire solution (that is attached along with the problem as a png file), except how to get [(x + Δx)^(4/3) + (x + Δx)^(2/3) x^(2/3) + x^(4/3)]/[(x + Δx)^(4/3) + (x + Δx)^(2/3) x^(2/3) + x^(4/3)] = 1, without looking at the solution.

    How would I come up with that on my own? What's “playing tricks with my head” is the fractional exponents. If it were dealing with square roots, for example, I would simply multiply the numerator and denominator by the conjugate of the numerator.

    Any input would be GREATLY appreciated!

    Attached Files:

  2. jcsd
  3. Dec 17, 2013 #2
    It's the cube root version of the conjugate multiplication trick; ##a-b=(a^{1/3}-b^{1/3})(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3})##. The motivation for this trick comes from the difference of cubes formula, or perhaps the more general ##x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...+xy^{n-2}+y^{n-1})##.
  4. Dec 17, 2013 #3


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    Sorry, I double-posted.
  5. Dec 17, 2013 #4


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  6. Dec 17, 2013 #5
    Well ... that equation is true, and I suppose it is relevant to this particular problem. That's not really the way that I look at it though. The way I see it, we're applying ##a-b=(a^{1/3}-b^{1/3})(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3})## where ##a=(x+\Delta x)^2## and ##b=x^2##. The fact that things are being squared really has nothing to do with the trick that's being used. It's just something that happens to show up in this particular instance and has (in my opinion) very little to do with the work that's being done.

    In my mind, your equation adds an extra parameter to the situation and shifts attention away from the actual idea that is being used. I'm not saying it's wrong to think about it that way, it's just not the way that I think about it. If that's what works for you, go for it.

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