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Math Help - Proving the Product Rule

  1. #1
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    Proving the Product Rule

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

    I want to prove the product rule:

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

    And in the same way, with some rearrangement.

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

    Thanks in advance.
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  2. #2
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    Re: Proving the Product Rule

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

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

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

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

    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)) =

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

    therefore:

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

    = \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)

     = 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 \Delta f \Delta g, so to finish the proof, one has to show that:

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

    so that one can conclude:

    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)

    = 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}

    = 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)
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  3. #3
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    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.
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    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
    \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 g must be continuous we have \lim _{\Delta  \to 0} g(x_0  + \Delta ) = g(x_0 ).

    That gives you the result at once.
    Last edited by Plato; January 28th 2013 at 10:11 AM.
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    Re: Proving the Product Rule

    Quote Originally Posted by Plato View Post
    You can also write is as
    \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 g must be continuous we have \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 \lim_{\Delta x \to 0} \frac{\Delta f \Delta g}{\Delta x} vanishes, since:

    \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)

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

    and recalling that:

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

    \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*

     = 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; January 28th 2013 at 10:18 AM.
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    Re: Proving the Product Rule

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

    \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)

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

    and recalling that:
    \Delta g = g(x_0 + \Delta x) - g(x_0)
    \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*
     = 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 \Delta g is defined.
    Whereas clearly \Delta x is defined as a variable, what is \Delta f~?
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  7. #7
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    Re: Proving the Product Rule

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

    \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).
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