Results 1 to 2 of 2

Math Help - Definition of the Derivative

  1. #1
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1

    Definition of the Derivative

    I am going over some of my notes, and I was taking a different way to get to the definition.

    Find the slopes between two points, x - h AND x + h. We want the "slope" or derivative at x, so set the limit as h approaches zero. How do I use algebra to go from this:

    lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}

    To this:


    lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    First we need a simple theorem.

    Theorem: Suppose g is a function defined on an open interval containing a (except possibly at a) so that \lim_{x\to a} g(x) = L where L is a real number. Suppose f is a function defined on an open interval containing L so that f is continous at L then f\circ g is defined in an open interval containing L (except possibly at L) and \lim_{x\to a}f(g(x)) = f(L).

    Now we define a derivative.

    Definition: Suppose f is a function defined on an open interval I containing a. Define the function g:I\setminus \{ a\} \mapsto \mathbb{R} as g(x) = \frac{f(x)-f(a)}{x-a} then g is defined on an open interval containing a (except at a). If it happens that \lim_{x\to a} g(x) exists (real number) then we say f is differenciable at a and we call this limit f'(a).

    Theorem 1: Suppose that f is differenciable at a defined on an open interval I containing a. Define the function g(x) = \frac{f(x+a) - f(a)}{x} for all x such that x+a\in I \mbox{ and }x\not = 0 then clearly g is defined on an open interval containing 0 (except at at 0) and \lim_{x\to 0}g(x) = f'(a).

    Proof: The part where g is defined around 0 is straightforward. Now to show \lim_{x\to 0}g(x) = f'(a) we use the theorem above. Let h(x) = x + a and let F(x) = \frac{f(x)-f(a)}{x-a} then the conditions of the theorem are satisfied and f'(a)= \lim_{x\to 0}F(h(x)) = \lim_{x\to a}\frac{f(x+a) - f(a)}{x}.

    Theorem 2: Suppose that f is differenciable at a defined on an open interval I containing a. Define the function g(x) = \frac{f(a) - f(x-a)}{x} for all x such that x-a\in I \mbox{ and }x\not = 0 then clearly g is defined on an open interval containing 0 (except at 0) and \lim_{x\to 0}g(x) = f'(a).

    Proof: The proof is nearly identical to Theorem 1.

    Corollarly: Suppose that f is differenciable at a defined on an open interval I containing a. Define the function g(x) = \frac{f(x+a) - f(x-a)}{2x} for all x such that x\pm a \in I \mbox{ and }x\not = 0 then clearly g is defined on an open interval containing 0 (except at 0) and \lim_{x\to 0}g(x) = f'(a).

    Proof: Note that \frac{f(x+a) - f(x-a)}{2x} = \frac{1}{2}\cdot \frac{f(x+a) - f(a)}{x} + \frac{1}{2} \cdot \frac{f(a) - f(x-a)}{x}. This means \lim_{x\to 0}\frac{f(x+a)-f(x-a)}{2x} = \frac{1}{2}f'(a) + \frac{1}{2}f'(a) = f'(a) by the above theorems.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using the definition of a derivative.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 31st 2010, 06:22 PM
  2. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 06:33 AM
  3. Replies: 2
    Last Post: September 1st 2010, 02:29 PM
  4. Definition of derivative
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 14th 2010, 09:30 PM
  5. Replies: 2
    Last Post: November 6th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum