Understanding Derivatives: Calculus Homework Statement and Proof Explanation

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Homework Help Overview

The discussion revolves around understanding the proof for derivatives, specifically focusing on the limit expression involving the difference of powers, lim x-> 0 [xm - am]/[xn - an] and its relation to (m/n)am-n. Participants are exploring the nuances of this limit within the context of calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants attempt to factor the expressions involved and question how the exponents behave during this process. Others express confusion about the equality of the limit and the presence of the (m/n) factor. There are discussions about specific cases of m and n, and participants are trying to understand the general form of the limit.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants have provided insights into the factoring process and the implications of different values for m and n. There is a recognition of potential errors in the original video reference, and participants are encouraged to clarify the problem statement for better understanding.

Contextual Notes

Participants note that the limit may need to be evaluated as x approaches a rather than 0, and there is a call for a complete problem statement to better frame the discussion. The complexity of the proof and the nature of the powers involved are also highlighted as factors contributing to the confusion.

robertjford80
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Homework Statement



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





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
 
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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.
 
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.
 
robertjford80 said:
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.

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?
 
robertjford80 said:

Homework Statement



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





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

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.

RGV
 
Dick said:
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?

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:
Ray Vickson said:
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.
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.
 
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.

Screenshot2012-05-22at73851PM.png
 
Last edited by a moderator:
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]

and

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.
 
  • #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.
 
  • #11
e^(i Pi)+1=0 said:
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]

and

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.

That's not what I need to know.
 
  • #12
gopher_p said:
(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.


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.
 
  • #13
robertjford80 said:
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.

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.

RGV
 
Last edited:
  • #14
Do you mean

[xm - am]/[xn - an ≠ (m/n)am-n ?
 

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