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Homework Help: Calculus; Derivatives

  1. May 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Ok, I'm trying to understand the proof for derivatives. I understand most of it, but there is one step that I cannot understand.

    lim x-> 0 [xm - am]/[xn - an = (m/n)am-n

    3. The attempt at a solution

    I don't see how those are equal. The best I can do is

    (x-a)m(x-a)-n = (x-a)m-n
  2. jcsd
  3. May 22, 2012 #2


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    Factor (x^m-a^m). One factor is (x-a). What's the other one? Do the same thing for the denominator.
  4. May 22, 2012 #3
    I still don't get it.

    The best I can do is

    (x-a)(x+a) for the num and denom but I can't figure out what happens to the exponents.
  5. May 22, 2012 #4


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    That's the special case m=2. You also have other cases [itex]x^3-a^3=(x-a)(x^2+xa+a^2)[/itex] and [itex]x^4-a^4=(x-a)(x^3+x^2a+a^2x+a^3)[/itex]. What does the general case [itex]x^m-a^m[/itex] look like?
  6. May 22, 2012 #5

    Ray Vickson

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    The reason it is hard to understand is that it is FALSE. We have
    [tex] \lim_{x \rightarrow 0} \frac{x^m - a^m}{x^n - a^n} = a^{m-n}.[/tex] There should be no factor (m/n) in front. Are you sure you are looking at the correct ratio or the correct limit? Maybe you don't want x --> 0.

  7. May 22, 2012 #6
    I don't know. I've already tried factoring and the best I could do is

    (x-a)m(x-a)-n = (x-a)m-n
    Last edited: May 22, 2012
  8. May 22, 2012 #7


    Staff: Mentor

    You need to address what Ray said. Since this is a derivative, the limit is probably as x approaches a, not 0. Aside from that, it's not obvious to me what the original function you're differentiating might be.

    Please provide the complete problem statement.
  9. May 22, 2012 #8
    It's a proof for the power rule of derivatives. I watched it in this video:

    Here's a screen shot, the handwriting is very bad.

    Last edited by a moderator: Sep 25, 2014
  10. May 22, 2012 #9
    There are two long-hand methods available to us here.

    1. [itex]\frac{df}{dx}= \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/itex]


    2. m[itex]_{tan}=\lim_{x \rightarrow a}[/itex][itex]\frac{f(x)-f(a)}{x-a}[/itex]

    You'll want to use the first to take the derivative and the second to directly find the slope of the tangent at point x.
  11. May 22, 2012 #10
    (1) ##x^n+y^n\neq(x+y)^n## unless either ##x=0## or ##y=0## or ##n=1##. This is a very common mistake among calculus students. Don't make this mistake. You will lose every time.

    (2) Understanding the proof of the power rule is not likely to help you to be able to use the power rule or improve your grade in a standard calculus class. I'm not trying to discourage you from doing so, in fact I applaud your effort. I'm just saying ...

    (3) Because it's easier to verify than it is to derive (even with some examples), ##x^k-a^k=(x-a)(x^{k-1}+x^{k-2}a+x^{k-3}a^2+x^{k-4}a^3+...+x^3a^{k-4}+x^2a^{k-3}+xa^{k-2}+a^{k-1})##. But this only works when ##k## is a natural number.

    (4) The guy in the video made a typo. It should have been ##lim_{x\rightarrow a}\frac{x^m-a^m}{x^n-a^n}=\frac{m}{n}a^{m-n}##. The first step in this limit is factoring as in (3). Again, we need ##m## and ##n## to be natural numbers.

    (5) To prove the power rule in general (i.e. for powers that aren't natural numbers) one typically appeals to a combination of one or more of (i) "easy" power rule, (ii) the chain rule, (iii) the quotient rule, (iv) implicit differentiation, (v) logarithmic differentiation.
  12. May 22, 2012 #11
    That's not what I need to know.
  13. May 22, 2012 #12

    I'm not taking a class and I've known how to do derivatives for about 2 months, now I'm finally getting around to understanding what derivatives are all about.

    If the guy in the video made a typo in the video then that's fine, I still don't know how he gets the m/n as a coefficient.
  14. May 23, 2012 #13

    Ray Vickson

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    You are wasting your time. The guy in the video made a mistake. He should not have gotten what he wrote. Trying to understand it will get you exactly nowhere. On the other hand, if you want to understand why
    [tex] \lim_{x \rightarrow a} \frac{x^n - a^n}{x^m - a^m} = \frac{n}{m}a^{n-m},[/tex] all you need to do is read the responses that have already been posted, showing you exactly why.

    Last edited: May 23, 2012
  15. May 23, 2012 #14
    Do you mean

    [xm - am]/[xn - an ≠ (m/n)am-n ?
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