Understanding Derivatives: Calculus Homework Statement and Proof Explanation

In summary, the conversation is about understanding the proof for the power rule of derivatives. The main point of confusion is about a step that involves taking the limit of a ratio. One person suggests using two long-hand methods to understand the proof, while another points out a common mistake in calculating powers and provides a method for verifying the result. The conversation also touches on the importance of understanding the general case and how to prove the power rule in general. Despite some confusion and a possible typo in the video, the person is determined to understand derivatives.
  • #1
robertjford80
388
0

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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  • #2
Factor (x^m-a^m). One factor is (x-a). What's the other one? Do the same thing for the denominator.
 
  • #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.
 
  • #4
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?
 
  • #5
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
 
  • #6
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:
  • #7
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.
 
  • #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.

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

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
 
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  • #14
Do you mean

[xm - am]/[xn - an ≠ (m/n)am-n ?
 

Related to Understanding Derivatives: Calculus Homework Statement and Proof Explanation

1. What is a derivative?

A derivative is a mathematical concept that represents the instantaneous rate of change of a function at a specific point. It is calculated through the use of calculus and is often used in applications such as physics, engineering, and economics.

2. How do you find the derivative of a function?

The derivative of a function can be found by using the rules of calculus, specifically the power rule, product rule, quotient rule, and chain rule. These rules involve taking the limit of a function as the change in the input (x-value) approaches zero.

3. What is the significance of the derivative?

The derivative is significant because it allows us to understand how a function changes at a specific point. It can also be used to find maximum and minimum points, determine the concavity of a function, and solve optimization problems.

4. How is the derivative related to the graph of a function?

The derivative is related to the graph of a function by representing the slope of the tangent line at a specific point on the graph. The derivative can also be used to find the slope of a secant line, which is an approximation of the tangent line.

5. Can you provide an example of a derivative calculation?

Sure, let's find the derivative of the function f(x) = 3x^2 + 2x + 5. Using the power rule, we can rewrite the function as f(x) = 3x^2 + 2x^1 + 5x^0. Taking the derivative, we get f'(x) = 6x + 2. This means that at any point on the graph of f(x), the slope of the tangent line is given by the equation 6x + 2.

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