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 $\displaystyle x_0 + \Deltax$ (x_{0} 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:

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

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

$\displaystyle = f(x_0)g'(x_0) + g(x_0)f'(x_0) + 0 = f(x_0)g'(x_0) + g(x_0)f'(x_0)$

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.

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

$\displaystyle \frac{{h(x_0 + \Delta ) - h(x_0 )}}{\Delta } = \frac{{f(x_0 + \Delta ) - f(x_0 )}}{\Delta }g(x_0 + \Delta ) + f(x_0 )\frac{{g(x_0 + \Delta ) - g(x_0 )}}{\Delta }$

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.

Re: Proving the Product Rule

Quote:

Originally Posted by

**Plato** 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 x_{0}, we cannot say that g(x_{0}+Δx) is anywhere near g(x_{0})).

EDIT: the Wikipedia page : Product rule - Wikipedia, the free encyclopedia has several different proofs, and some nice pictures, too.

Re: Proving the Product Rule

Quote:

Originally Posted by

**Deveno** 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 x_{0}, we cannot say that g(x_{0}+Δx) is anywhere near g(x_{0})).

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 $\displaystyle \Delta f~?$

Re: Proving the Product Rule

Quote:

Originally Posted by

**Plato** 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, x_{0} plays the role of a constant).