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Finding the derivitive using limits?

  1. Sep 29, 2008 #1
    I've stumbled across a question. It must be solved using limits. I just don't know where to start. After doing the f(x+h) - f (x) / h i end up with a cluster that cannot be simplified... any help would be great.

    the equation is as follows.

    y = cube root( X^4 - 3X)
  2. jcsd
  3. Sep 30, 2008 #2
    Very interesting. I'm not sure how to prove this either, but I'm really interested in seeing the solution!

    I think the core difficulty in this problem is with the root, so let's simplify the problem. How do we find f' for f(x) = sqrt(x)?

    Then, we can probably generalize to n-th roots and use the chain rule to finish it off.

    My guess is that you have to abuse the relation [tex]\sqrt[n]{x} = e^{\ln(x^{\frac{1}{n}})} = e^{n \ln(x)}[/tex]. If you were a brutish proof person, you could probably do something with the taylor series of [tex]e^x[/tex] and ln(x) to solve it with limits.
    Last edited: Sep 30, 2008
  4. Sep 30, 2008 #3


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    The difference of two cubes can be factored this way:

    A^3 - B^3 = (A-B)(A^2 + AB + B^2)

    This works for any quantities [tex] A, B[/tex], so (curse the database errors which prevent preview from working)

    a-b = \left(a^{1/3} - b^{1/3}\right)\left(a^{2/3} + a^{1/3} b^{1/3} + b^{1/3}\right)

    The numerator of the difference quotient has

    a & = \left((x+h)^4 - 3(x+h)\right)^{1/3}\\
    b & = \left(x^4 - 3x \right)^{1/3}

    To rationalize the difference quotient multiply by this fraction:

    \frac{\left((x+h)^4 - 3(x+h)\right)^{2/3} + \left((x+h)^4 - 3(x+h)\right)^{1/3} \left(x^4 - 3x\right)^{1/3} + \left(x^4 - 3x\right)^{2/3}}{\left((x+h)^4 - 3(x+h)\right)^{2/3} + \left((x+h)^4 - 3(x+h)\right)^{1/3} \left(x^4 - 3x\right)^{1/3} + \left(x^4 - 3x\right)^{2/3}}

    this will leave you with (use the second factoring formula from the top of my post)

    \frac{\left((x+h)^4 - 3(x+h)) - (x^4 - 3x)\right)}{h\left((x+h)^4 - 3(x+h)\right)^{2/3} + \left((x+h)^4 - 3(x+h)\right)^{1/3} \left(x^4 - 3x\right)^{1/3} + \left(x^4 - 3x\right)^{2/3}}

    This still isn't pretty, but the work of taking the limit can now be performed.
  5. Sep 30, 2008 #4
    whoa thanks... now trying to understand it.

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