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A question about finding the derivative

  1. Apr 29, 2014 #1

    adjacent

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    Finally I've learned what a derivative is!
    I've just started self-learning calculus.It's really interesting.It will be taught next year anyway.
    I just want to solidify my understanding before officially studying it.
    If I am asked to find the derivative of ##f(x)=x^2##.What I do is select two points and find the slope of the line passing throught two points.
    $$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

    $$=\frac{(x+\Delta x)^2-x^2}{\Delta x}$$
    So the derivative of the graph at ##(x,f(x))## is the limit of the ##\frac{(x+\Delta x)^2-x^2}{\Delta x}## as ##\Delta x## approaches zero.
    But if I make ##\Delta x## zero here,the fraction tuns to ##0 \over 0## :eek:

    Then if I simplify it to ##\frac{\Delta x(2x+ \Delta x)}{\Delta x}##, I get 2x as the derivative.
    Why? Isn't ##\frac{(x+\Delta x)^2-x^2}{\Delta x}=\frac{\Delta x(2x+ \Delta x)}{\Delta x}##
    This is my problem.I don't have an intuitive understanding of this.
     
  2. jcsd
  3. Apr 29, 2014 #2

    Fredrik

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    That's why we take the limit instead of simply setting ##\Delta x## to 0.

    We have
    $$\frac{(x+\Delta x)^2-x^2}{\Delta x} = \Delta x+2x$$ But this isn't the derivative. It's an approximation of the derivative that gets better the smaller you make ##\Delta x##. The derivative is what you get from the above when you take the limit ##\Delta x\to 0##.
     
  4. Apr 29, 2014 #3

    adjacent

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    I guess I will need to learn more on limits.
    I'll come back after learning it.
     
  5. Apr 29, 2014 #4

    adjacent

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    I read from the lecture notes that we should consider ##\Delta x \neq 0## . If so,why do we make x zero at last?
     
  6. Apr 29, 2014 #5

    micromass

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    We never make ##\Delta x## zero. The limit says what the expression will be very close to provided that ##\Delta x## is very close but unequal to zero. So if we say

    [tex]\lim_{\Delta x\rightarrow 0} f(\Delta x) = c[/tex]

    then we mean that if ##\Delta x## is very close to ##0##, then ##f(\Delta x)## is very close to ##c##. We never actually take ##f(0)##.

    So in your example, we never get ##0/0## since ##\Delta x## is allowed to be very close to ##0## but unequal to ##0##.

    Of course, you have probably seen and applied the following identity:

    [tex]\lim_{\Delta x\rightarrow 0} f(\Delta x) = f(0)[/tex]

    this seems to contradict what I said about evluating ##f(0)##. But note that this identity is only true in very special cases. It can only be used when ##f## is continuous and well-defined in ##0##. So if you have

    [tex]\lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}[/tex]

    then you cannot substitute ##0## in the fraction because the fraction is not well-defined in ##0##. So the above rule does not apply. However, once you simplify it to

    [tex]\lim_{\Delta x\rightarrow 0} \Delta x + 2x[/tex]

    Then you can apply the rule since the function is continuous and well-defined in ##0##. So the answer is ##2x##.
     
  7. Apr 29, 2014 #6

    Fredrik

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    Our goal is actually to show that for each positive real number ##\varepsilon##, there's a positive real number ##\delta## such that for all non-zero values of ##\Delta x## between ##-\delta## and ##\delta##, we have
    $$ \left|\frac{(x+\Delta x)^2-x^2}{\Delta x}-2x\right|<\varepsilon.$$ This is what it means to say that
    $$\frac{(x+\Delta x)^2-x^2}{\Delta x}\to 2x\quad\text{as }\Delta x\to 0.$$

    Our calculation shows that the choice ##\delta=\varepsilon## will get the job done (for this specific function):

    Let ##\varepsilon>0## be arbitrary. Define ##\delta=\varepsilon##. For all non-zero ##\Delta x## such that ##|\Delta x|<\delta##, we have
    $$ \left|\frac{(x+\Delta x)^2-x^2}{\Delta x}-2x\right|=|\Delta x+2x-2x|=|\Delta x|<\varepsilon.$$

    When you use a calculation like
    $$\frac{(x+\Delta x)^2-x^2}{\Delta x}=\Delta x+2x\to 2x$$ to determine that the limit is 2x, you're relying on some theorems about limits of functions. If you haven't worked through the proofs of those theorems, you won't fully understand why the method works. If you haven't even read the statements of those theorems, it's almost impossible to understand calculations like this.

    (I wrote half of this before micromass wrote his post. I put it aside when I saw that he had already said most of what I wanted to say. I came back after a while and decided to elaborate a bit).
     
  8. Apr 30, 2014 #7

    adjacent

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    Where can I read these theorems?
    and from where did that ##-2x## come from?
     
    Last edited: Apr 30, 2014
  9. Apr 30, 2014 #8

    Fredrik

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    Any book with "calculus" or "analysis" in the title should cover them (but you would probably find the ones with "analysis" in the title too hard). There are also places like Wikipedia and MathWorld. And, I haven't even looked at it myself, but people say that Khan Academy is very nice.

    You should probably try to understand limits of sequences before you try to understand limits of functions. The definition is similar, but easier.

    The -2x is in there because that calculation isn't supposed to find the limit. It's supposed to prove that the limit is 2x. It's the calculation you might want to do once you have found some other reason to think that the limit is 2x. If we want to prove e.g. that ##g(h)\to 2x## as ##h\to 0##, the proof involves showing that g(h) is "close enough" to 2x when h is small enough. To say that g(h) is "close" to 2x is to say that |g(h)-2x| is small.

    What exactly does this "close enough"/"small enough" stuff mean? That's exactly what the definition of "limit" (of a function) tells you.

    Edit: The videos near the bottom of that Khan academy page, in the "epsilon delta" section should be a good place to start. Direct link to the first one.
     
    Last edited: Apr 30, 2014
  10. Apr 30, 2014 #9

    adjacent

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    Thank you.

    I guess my knowledge on limits is still not that solid.

    Thanks Fredrik and micromass!

    I'll come back after sometime.(I mean after studying a calculus book)
    It's kinda odd to study alone. :(. I never did this before but calculus is really interesting so I couldn't resist. :wink:
     
  11. Apr 30, 2014 #10

    micromass

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    The book "First Course in Calculus" by Lang is very nice. It states all the relevant theorems, but sadly enough does not prove them until one of the last chapters (and the chapter isn't that very well written).
    Books like Apostol and Spivak however should contain everything you want to know, but they are quite difficult for beginners.
     
  12. Apr 30, 2014 #11

    statdad

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    Yes
    [tex] \frac{(x+\Delta x)^2 - x^2}{\Delta x} = \frac{\Delta x\left(2x + \Delta x\right)}{\Delta x}
    [/tex]
    The reason you end up with 0/0 when you set [itex] \Delta x = 0 [/itex] in the expression on the left is that there is a common factor of [itex] \Delta x [/itex] in numerator and denominator. That fact is hidden in the form on the left, explicit in the form on the right.
    The rest of the limit process works because in the simplification the common factor is cancelled.
     
  13. Apr 30, 2014 #12

    adjacent

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    Thanks guys.
     
  14. Apr 30, 2014 #13

    Ray Vickson

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    You are struggling with concepts that give more students trouble than any others, and which are very easy to get wrong. So, do not feel bad if you are having difficulties grasping the material; that is its nature!

    Throughout much of the history of calculus the rigorous concept of limit was not used and perhaps did not even 'exist' in the current books of knowledge then available. For example, Newton, Liebnitz, Gauss, Laplace, Euler and others made huge advances in calculus knowledge (developing perhaps most of what we use in applications, even today) but without using limits in any significant way. Even some 19th century "Mathematics for Physics" type books use "infinitesimals" more than limits, if they use limits at all. However, it has been known for a long time that the 'infinitesimals' that people like Leibnitz used are not well-founded, so mathematicians ousted their use and re-wrote everything in terms of limits.

    Surprisingly, in the late 20th Century the concepts and uses of infinitesimals has been re-established in a now rigorous way (see 'Non-Standard Analysis'), and many of the results obtained using standard calculus have be re-derived using this alternative approach---but which is still not widely embraced by mathematicians as a whole. There have even been attempts to teach calculus using the (modern) concept of infinitesimals, so by-passing the whole business of limits altogether. You can see one such attempt in the (free) elementary calculus textbook by Keisler, available as a pdf file at http://www.math.wisc.edu/~keisler/calc.html . To quote from the preface of the first Edition: "The calculus was originally developed using the intuitive concept of an infinitesimal, or an infinitely small number. But for the past one hundred years infinitesimals have been banished from the calculus course for reasons of mathematical rigor. Students have had to learn the subject without the original intuition. ... "

    I am not sure whether my mentioning this stuff to you is doing you a service or a dis-service, but anyway, there it is.
     
  15. Apr 30, 2014 #14

    adjacent

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    That helped for sure.
    A concept can only be understood well if you know how it came to life.Well,that's how it is for me!

    btw,that book seems to be very good.It's downloading now.
     
  16. Apr 30, 2014 #15

    lurflurf

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    I'll just restate what has been said in my own words.

    A method commonly used, but poorly explained in calculus books is passage to the continuous extension. We can find the limit of a continuous function by evaluation, if the function is not continuous we can (when the limit exists) replace it with a continuous fuction and find the limit by evaluation of the new function.

    $$\lim_{x\rightarrow a}\mathrm{f}(x)=\bar{\mathrm{f}}(a)$$

    In practice we do this by finding a continuous function that equals the given function near the given point.

    $$\frac{(x+\Delta x)^2-x^2}{\Delta x}\underset{\Delta x \ne 0}=2x+\Delta x$$

    The confusing thing is that limits do not depend on the value at the point in question, but can be found from the value for a continuous function.
     
    Last edited: Apr 30, 2014
  17. May 1, 2014 #16

    adjacent

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    Thanks lurf,
    btw, do you think these video lectures are good for a beginner like me?
     
  18. May 1, 2014 #17

    Fredrik

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    I took a quick look at the beginning of video 2, titled "limits". It looks good, but it's actually about derivatives. My recommendation is that you first learn about limits of sequences*, then limits of functions (the Khan academy videos), and then watch a few of these videos.

    *) If you make sure that you understand the definition of limits of sequences, it will be a nice "warm up" to the more difficult definition of limits of functions.
     
  19. May 1, 2014 #18

    adjacent

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    Thanks for the suggestion!
    Can you suggest some video lectures about limits of sequences?*
    I found the wikipedia article difficult to comprehend.

    *I find video lectures easier to understand than books or texts.
     
  20. May 1, 2014 #19

    Fredrik

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    Actually the definition of limits of sequences is simple enough that I can explain it right here. I will use the notation (a,b) for the "open interval" from a to b, i.e. the set of all real numbers x such that a < x < b. The word "open" refers to the fact that the endpoints aren't included in the interval. For each x and each r>0, the open interval (x-r,x+r) has x right at the center. I will call this the open interval of radius r centered at x.

    Let ##x_1,x_2,x_3,...## be a sequence of real numbers. The numbers ##x_1##, ##x_2## and so on are called the terms of the sequence. I use the notation ##\langle x_n\rangle_{n=1}^\infty## for the sequence itself. The notations ##(x_n)_{n=1}^\infty## and ##\{ x_n\}_{n=1}^\infty## are also used. I think the first one is fine, but I strongly disapprove of that last one. In my opinion it should be used for the (unordered) set of all terms, but not for the sequence itself.

    Here's the definition:
    A real number x is said to be a limit of ##\langle x_n\rangle_{n=1}^\infty## if every open interval centered at x contains all but a finite number of terms of the sequence. ​
    Let say that in a slightly different way:
    A real number x is said to be a limit of ##\langle x_n\rangle_{n=1}^\infty## if for all positive real numbers r, the number of terms that are not in the interval (x-r,x+r) is finite.​
    The most common way of stating the definition is a bit more complicated. I guess people still prefer it because they feel that it's not 100% clear what phrases like "all but a finite number" means. It goes like this:
    A real number x is said to be a limit of ##\langle x_n\rangle_{n=1}^\infty## if for all ##\varepsilon>0##, there's a positive integer N such that for all integers n, if ##n\geq N## then ##|x_n-x|<\varepsilon##.​
    It's traditional to use the Greek letter ##\varepsilon## (epsilon) for the "arbitrarily small but stilll positive" real number in these definitions.

    The statement "x is a limit of ##\langle x_n\rangle_{n=1}^\infty##" is written as "##x_n\to x## as ##n\to\infty##", or as ##\lim_{n\to\infty} x_n=x##. (The latter notation would not make sense if a sequence could have two different limits, but we can prove that it doesn't).

    Here's how to use the definition to show that 0 is a limit of ##\langle \frac 1 n\rangle_{n=1}^\infty##. Let ##\varepsilon## be an arbitrary positive real number. Let N be an integer such that ##N\varepsilon>1##. For all integers n such that ##n\geq N##, we have
    $$\left|\frac 1 n - 0\right|=\frac 1 n \leq \frac 1 N < \varepsilon.$$
     
    Last edited: May 1, 2014
  21. May 1, 2014 #20

    adjacent

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    I don't understand this part.Form where did r come from?

    And can you please explain what this is?
    If only I understand this,will I be able to understand the other parts :redface:
     
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