Results 1 to 7 of 7

Thread: Proving the Product Rule

  1. #1
    Junior Member
    Joined
    Apr 2012
    From
    South Africa
    Posts
    42
    Thanks
    3

    Proving the Product Rule

    So I'm having a bit of trouble here...

    I want to prove the product rule:

    $\displaystyle \displaystyle \begin{align*} h(x) = f(x)g(x) \\ h'(x) = f'(x)g(x) + f(x)g'(x) \\ \end{align*}$

    So assuming that f and g are differentiable at Xo:

    $\displaystyle \displaystyle \begin{align*} \Delta h=h(x_0+\Delta x)-h(\Delta x)\\ \\ \end{align*}$

    And in the same way, with some rearrangement.

    $\displaystyle \displaystyle \begin{align*} \Delta f=f(x_0+\Delta x)-f(\Delta x)\\ f(x_0+\Delta x) = \Delta f + f(\Delta x) \rightarrow Eqn. 1 \\ \\ \Delta g=g(x_0+\Delta x)-g(\Delta x)\\ g(x_0+\Delta x) = \Delta g + g(\Delta x) \rightarrow Eqn. 2\end{align*}$

    And, obviously:

    $\displaystyle \displaystyle \begin{align*} h(x_0+\Delta x) = f(x_0+\Delta x)\times g(x_0+\Delta x) \end{align*}$

    Substituting equations 1 and 2 and multiplying out yields:

    $\displaystyle \displaystyle \begin{align*} h(x_0+\Delta x)=\Delta f \Delta g + f(x_0)\Delta g+g(x_0)\Delta f \end{align*}$

    So now the example I'm following just leaves things here, so apparently this is enough, but I don't see how this simplifies to what I understand to be the goal of the exercise i.e. $\displaystyle \displaystyle \begin{align*} \Delta h= f(x)\Delta g+g(x)\Delta f \end{align*}$

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    Re: Proving the Product Rule

    first of all, i am confused by the definition of $\displaystyle \Delta h$. as i see it, this should be:

    $\displaystyle \Delta h = h(x_0 + \Delta x) - h(x_0)$

    that is, we want $\displaystyle \Delta h$ to represent the change in h between the points $\displaystyle x_0$ and $\displaystyle x_0 + \Deltax$ (x0 plus "a little bit more (plus/minus)"). what you have doesn't look right at ALL, since $\displaystyle \Delta h \to h(x_0) - h(0)$ as $\displaystyle \Delta x \to 0$, but the idea should be: "h changes very little, if x does".

    writing similar definitions for $\displaystyle \Delta f,\Delta g$:

    $\displaystyle h(x_0 + \Delta x) = f(x_0 + \Delta x)\cdot g(x_0 + \Delta x) = (\Delta f + f(x_0))(\Delta g + g(x_0)) =$

    $\displaystyle \Delta f \Delta g + f(x_0)\Delta g + g(x_0)\Delta f + f(x_0)g(x_0)$

    therefore:

    $\displaystyle \Delta h = h(x_0 + \Delta x) - h(x_0) $

    $\displaystyle = \Delta f\Delta g + f(x_0)\Delta g + g(x_0)\Delta f + f(x_0)g(x_0) - f(x_0)g(x_0)$

    $\displaystyle = f(x_0)\Delta g + g(x_0)\Delta f + \Delta f\Delta g$

    this is almost what we want, the only "sticking point" is the term $\displaystyle \Delta f \Delta g$, so to finish the proof, one has to show that:

    $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f\Delta g}{\Delta x} = 0$

    so that one can conclude:

    $\displaystyle h'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta h}{\Delta x} = \lim_{\Delta x \to 0}\left( f(x_0)\frac{\Delta g}{\Delta x} + g(x_0)\frac{\Delta f}{\Delta x} + \frac{\Delta f\Delta g}{\Delta x}\right)$

    $\displaystyle = f(x_0)\lim_{\Delta x \to 0}\frac{\Delta g}{\Delta x} + g(x_0)\lim_{\Delta x \to 0}\frac{\Delta f}{\Delta x} + \lim_{\Delta x \to 0} \frac{\Delta f\Delta g}{\Delta x}$

    $\displaystyle = f(x_0)g'(x_0) + g(x_0)f'(x_0) + 0 = f(x_0)g'(x_0) + g(x_0)f'(x_0)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2012
    From
    South Africa
    Posts
    42
    Thanks
    3

    Re: Proving the Product Rule

    I'm sorry! That was a really silly mistake on my part. Embarrassing stuff. I really appreciate you taking the time to correct me.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,734
    Thanks
    2810
    Awards
    1

    Re: Proving the Product Rule

    Quote Originally Posted by Ivanator27 View Post
    I'm sorry! That was a really silly mistake on my part. Embarrassing stuff. I really appreciate you taking the time to correct me.

    You can also write it as
    $\displaystyle \frac{{h(x_0 + \Delta ) - h(x_0 )}}{\Delta } = \frac{{f(x_0 + \Delta ) - f(x_0 )}}{\Delta }g(x_0 + \Delta ) + f(x_0 )\frac{{g(x_0 + \Delta ) - g(x_0 )}}{\Delta }$

    Because $\displaystyle g$ must be continuous we have $\displaystyle \lim _{\Delta \to 0} g(x_0 + \Delta ) = g(x_0 )$.

    That gives you the result at once.
    Last edited by Plato; Jan 28th 2013 at 10:11 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    Re: Proving the Product Rule

    Quote Originally Posted by Plato View Post
    You can also write is as
    $\displaystyle \frac{{h(x_0 + \Delta ) - h(x_0 )}}{\Delta } = \frac{{f(x_0 + \Delta ) - f(x_0 )}}{\Delta }g(x_0 + \Delta ) + f(x_0 )\frac{{g(x_0 + \Delta ) - g(x_0 )}}{\Delta }$

    Because $\displaystyle g$ must be continuous we have $\displaystyle \lim _{\Delta \to 0} g(x_0 + \Delta ) = g(x_0 )$.

    That gives you the result at once.
    the continuity of g is also why the term $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f \Delta g}{\Delta x}$ vanishes, since:

    $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f\Delta g}{\Delta x} = \left(\lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}\right)(\lim_{\Delta x \to 0} \Delta g)$

    $\displaystyle = f'(x_0)(\lim_{\Delta x \to 0} \Delta g)$

    and recalling that:

    $\displaystyle \Delta g = g(x_0 + \Delta x) - g(x_0)$

    $\displaystyle \lim_{\Delta x \to 0} \Delta g = (\lim_{\Delta x \to 0} g(x_0 + \Delta x)) - (\lim_{\Delta x \to 0} g(x_0))$

    and *because g is continuous*

    $\displaystyle = g(x_0) - g(x_0) = 0$

    (the continuity of g is essential, if g suddenly "jumps" at x0, we cannot say that g(x0+Δx) is anywhere near g(x0)).


    EDIT: the Wikipedia page : Product rule - Wikipedia, the free encyclopedia has several different proofs, and some nice pictures, too.
    Last edited by Deveno; Jan 28th 2013 at 10:18 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,734
    Thanks
    2810
    Awards
    1

    Re: Proving the Product Rule

    Quote Originally Posted by Deveno View Post
    the continuity of g is also why the term $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f \Delta g}{\Delta x}$ vanishes, since:

    $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta f\Delta g}{\Delta x} = \left(\lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}\right)(\lim_{\Delta x \to 0} \Delta g)$

    $\displaystyle = f'(x_0)(\lim_{\Delta x \to 0} \Delta g)$

    and recalling that:
    $\displaystyle \Delta g = g(x_0 + \Delta x) - g(x_0)$
    $\displaystyle \lim_{\Delta x \to 0} \Delta g = (\lim_{\Delta x \to 0} g(x_0 + \Delta x)) - (\lim_{\Delta x \to 0} g(x_0))$
    and *because g is continuous*
    $\displaystyle = g(x_0) - g(x_0) = 0$
    (the continuity of g is essential, if g suddenly "jumps" at x0, we cannot say that g(x0+Δx) is anywhere near g(x0)).
    My problem with that is simple: I have no idea how $\displaystyle \Delta g$ is defined.
    Whereas clearly $\displaystyle \Delta x$ is defined as a variable, what is $\displaystyle \Delta f~?$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    Re: Proving the Product Rule

    Quote Originally Posted by Plato View Post
    My problem with that is simple: I have no idea how $\displaystyle \Delta g$ is defined.
    Whereas clearly $\displaystyle \Delta x$ is defined as a variable, what is $\displaystyle \Delta f~?$
    as indicated in post 2, for any function f:

    $\displaystyle \Delta f = f(x_0 + \Delta x) - f(x_0)$

    this is a function of the variable Δx (in it, x0 plays the role of a constant).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Dec 20th 2011, 04:57 AM
  2. [SOLVED] quick question on product rule and equality rule for logs
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: Oct 19th 2011, 07:29 PM
  3. Replies: 9
    Last Post: Nov 9th 2010, 01:40 AM
  4. Replies: 5
    Last Post: Oct 19th 2009, 01:04 PM
  5. Replies: 3
    Last Post: May 25th 2009, 06:15 AM

Search Tags


/mathhelpforum @mathhelpforum