# Proving the Product Rule

• Jan 28th 2013, 07:49 AM
Ivanator27
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*}

• Jan 28th 2013, 09:04 AM
Deveno
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:

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

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

$= f(x_0)g'(x_0) + g(x_0)f'(x_0) + 0 = f(x_0)g'(x_0) + g(x_0)f'(x_0)$
• Jan 28th 2013, 09:23 AM
Ivanator27
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.
• Jan 28th 2013, 09:47 AM
Plato
Re: Proving the Product Rule
Quote:

Originally Posted by Ivanator27
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.
• Jan 28th 2013, 10:11 AM
Deveno
Re: Proving the Product Rule
Quote:

Originally Posted by Plato
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.
• Jan 28th 2013, 10:18 AM
Plato
Re: Proving the Product Rule
Quote:

Originally Posted by Deveno
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 $\Delta f~?$
• Jan 28th 2013, 10:21 AM
Deveno
Re: Proving the Product Rule
Quote:

Originally Posted by Plato
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).