Results 1 to 4 of 4

Math Help - k constant in first principle formula

  1. #1
    Banned
    Joined
    Apr 2010
    Posts
    58

    k constant in first principle formula

    Let f be a differentiable function. For any constant k > 0, put g(x) = f(kx). Prove from first principles that g'(a) = kf'(ka) for all a. Can you explain what this result means geometrically?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    I presume that by "first principles" you mean using g'(a)= \lim_{h\to 0}\frac{g(a+h)- g(a)}{h}.

    Just replace g(a) by kf(a).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Apr 2010
    Posts
    58
    I got:

    lim
    h -> 0 kf(a + h) - kf(a) / h

    lim
    h -> 0 k[f(a + h) - kf(a)] / h

    I dont see a way to eliminate the h
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    Quote Originally Posted by TsAmE View Post
    I got:

    lim
    h -> 0 kf(a + h) - kf(a) / h

    lim
    h -> 0 k[f(a + h) - kf(a)] / h

    I dont see a way to eliminate the h
    You don't! That is
    k\left(\lim_{h\to 0}\frac{f(a+h)- f(a)}{h}\right).

    Now, what is that equal to?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: July 15th 2011, 01:30 PM
  2. Gravitational constant formula
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: February 9th 2011, 05:11 AM
  3. Well ordering principle and the maximum principle
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: August 3rd 2010, 08:31 AM
  4. Momentum Principle and Energy Principle Problem
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: October 4th 2009, 01:42 AM
  5. Replies: 5
    Last Post: February 29th 2008, 02:05 PM

Search Tags


/mathhelpforum @mathhelpforum