Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - derivatives

  1. #1
    Newbie
    Joined
    Dec 2005
    Posts
    12

    Exclamation derivatives

    my homework..
    Last edited by shirel; August 28th 2007 at 12:04 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jones's Avatar
    Joined
    Apr 2006
    From
    Norway
    Posts
    170
    There are rules/formulae for these these kinds of operations.

    The addition formula, the product rule etc...
    Just use them.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jones View Post
    There are rules/formulae for these these kinds of operations.

    The addition formula, the product rule etc...
    Just use them.
    He is being asked to prove them

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by shirel View Post
    my homework..

    1. consider the sum of two functions s(x): =f(x) + g(x) as a new function. calculate the derivative of s(x) and prove thereby that it equals the sum of the derivatives of g and g.
    Consider:

    <br />
\frac{s(x+h)-s(x)}{h}=\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}<br />

    <br />
=\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}<br />

    So:

    <br />
\lim_{h \to 0}\frac{s(x+h)-s(x)}{h}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}<br />
+ \lim_{h \to 0}\frac{g(x+h)-g(x)}{h}<br />

    which if all of these limits exist proves that:

    s'(x)=f'(x)+g'(x).

    The others are similar, just use the definition of what a derivative is.
    Replace the function by its other representation and rearrange to the
    form we know it must have.


    RonL

    The meaning of this will become clear (as will my TeX errors) when the
    LaTeX system is reinstalled
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608
    Hello, shirel!

    #3 asks us to derive the Product Rule.
    . . Too bad that Latex is off-line right now.


    3. Consider the product of two functions p(x) = f(x)\cdot g(x)
    Calculate the derivative of p(x).

    We will derive: . p'(x)\:=\:\lim_{h\to0}\frac{p(x+h) - p(x)}{h}

    We have: . [(x+h) - p(x)\;=\;f(x+h)\cdot g(x+h) - f(x)\cdot g(x)


    Subtract and add f(x+h)\cdot g(x):

    p(x+h) - p(x)\;=\;f(x+h)\cdot g(x+h) - f(x+h)\cdot g(x) - f(x)\cdot g(x) + f(x+h)\cdot g(x)

    p(x+h) - p(x)\;=\;f(x+h)\left[g(x+h) - g(x)\right] + g(x)\left[f(x+h) - f(x)\right]


    Divide by h:

    [tex]\frac{p(x+h) - p(x)}{h} \;= \;f(x+h)\left[\frac{g(x+h)-g(x)}{h}\right] + g(x)\left[\frac{f(x+h)-f(x)}{h}\right] [/ math]


    Take the limit:

    p'(x)\;=\;\lim_{h\to0}f(x+h)\cdot\lim_{h\to0}\left[\frac{g)x+h) - g(x)}{h}\right] + g(x)\cdot\lim_{h\to0}\left[\frac{f(x+h)-f(x)}{h}\right]


    Therefore: . p'(x)\;=\;f(x)\cdot g'(x) + g(x)\cdot f'(x)

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    There is just one problem Soroban.
    Unlike, CaptainBlank who states that the limits exists and thus does their sum, you did not. In front of mathematicians your proof would not be acceptable. (This is the major difference between Calculus and Analysis).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608
    Hello, TPHacker!

    Silly me . . . I thought this was a Calculus I problem.

    I wasn't aware that this was a grad-school Advanced Analysis problem
    . . where Existence must be established first.

    I won't make that mistake again . . .

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Soroban View Post
    Hello, TPHacker!

    Silly me . . . I thought this was a Calculus I problem.
    I do not know if you are joking or not but there is no way this can be a Calculus I problem. The professor is not going to make you prove the product rule, which Leibniz was struggling to prove. I assume the students were familiar with the infromal proof from Calculus I but wanted them to show existence of limits. (Honestly how many people can spot that add-subtract trick).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Oh, come Perfect! It was proved in very Calculus I class I ever taught. It is a standard way of illustrating that the ‘derivative limit’ does not have to be written in the order we all are used to seeing.

    BTW: It usually goes, “There are 10 kinds of people: those who understand binary numbers and those who don’t.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Plato View Post
    Oh, come Perfect!
    Come where?

    Quote Originally Posted by Plato
    It was proved in very Calculus I class I ever taught.
    Maybe, you taught an advanced class. Because I realized teachers do not like to get too theoretic about limits, they treat them more intuitevly.

    I understand what you are saying that the proof is not so difficult, yes, I agree. I am saying that you need to show the two functions are differenciable and use rules about the existence of limits otherwise the proof is not sound.

    Quote Originally Posted by Plato
    BTW: It usually goes, “There are 10 kinds of people: those who understand binary numbers and those who don’t.
    I know, that is the joke.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,838
    Thanks
    320
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Maybe, you taught an advanced class. Because I realized teachers do not like to get too theoretic about limits, they treat them more intuitevly.

    I understand what you are saying that the proof is not so difficult, yes, I agree. I am saying that you need to show the two functions are differenciable and use rules about the existence of limits otherwise the proof is not sound.
    As it happens, when I did my undergrad proofs like these were usually saved for the "Advanced Calculus" class in the fourth semester.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by topsquark View Post
    As it happens, when I did my undergrad proofs like these were usually saved for the "Advanced Calculus" class in the fourth semester.
    But in an advanced calculus class I would hope that you gave epsilon-delta proof for these properties. Not simple limit proofs. Now there I do agree: epsilon-delta proofs have no place in beginning calculus.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,838
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Plato View Post
    But in an advanced calculus class I would hope that you gave epsilon-delta proof for these properties. Not simple limit proofs. Now there I do agree: epsilon-delta proofs have no place in beginning calculus.
    Actually, I never took the class so I can't say for sure. But I believe you are right about it.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Plato View Post
    But in an advanced calculus class I would hope that you gave epsilon-delta proof for these properties. Not simple limit proofs. Now there I do agree: epsilon-delta proofs have no place in beginning calculus.
    Once you have extablished the basic rules of limits through delta-epsilon you can stop using it for more advanced limit proofs.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Once you have extablished the basic rules of limits through delta-epsilon you can stop using it for more advanced limit proofs.
    Not sure what you point is? But I can tell you that over a thirty year stint of teaching real analysis, undergraduate and graduate, we never used anything but the epsilon-delta proofs. Limits, continuity, derivatives, and integrals are all defined using epsilon-deltas. We insist, well at least the school of mathematics I represent, a solid proof using epsilon and deltas is the only correct proof.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives with both a and y
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2009, 09:17 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum