Talk:Product rule

From Wikipedia, the free encyclopedia

Hi,

the proof given for the product rule is a little hard to follow. What about rewriteing it, using the x -> x0 notation? And explaining the steps in more depth (I've had a little difficulty proving the tricky part to myself, when g(x)h(x) - g(x0)h(x0) becomes g(x) (h(x) - h(x0)) + h(x) ( g(x) - g(x0)).) PAStheLoD 18:49, 9 June 2006 (UTC)

---

Thanks, Anome. But if we have separate "formal" proofs that:

a) As δx -> 0, δy/δx -> dy/dx (what I have called "differentiation from first principles")
b) The limit of a product is the product of the limits, and vice versa (what I have called the "product rule for limits")

then does that make the proof count as a formal one? Kidburla2002.

The justification is informal, because it uses infinitesimals in an informal way: it works under most reasonable assumptions, but what about unreasonable ones? To do it formally, you can either use limits formally (with δ - ε type stuff), or a formalised system of infinitesimals. The Anome


Why was this moved? What is going at Product rule? -- Tarquin

I am not aware of any other product rule, so a disambiguation like product rule (calculus) is unnecessary and I will move back to product rule. AxelBoldt 19:54 18 Jun 2003 (UTC)

The Product Rule is a proper name Pizza Puzzle

There's plenty of other identities which can be called product rules. There must be 15 of them involving curls and divs and grads and what not. But I'm an opponent of pre-emptive disambiguation, so I say leave it unless there's a real need to move it. Certainly you can call this one the product rule, and get away with it. -- Tim Starling 14:24 19 Jun 2003 (UTC)

There is another "product rule" that is often used in combinatorics. (see Rule of product) which (i think) would be better listed as "product rule" than "rule of product". --Hughitt1 23:09, 3 April 2006 (UTC)

I took away the "informal justification" section, for the following reasons:

  1. It is virtually identical in essence to the argument using differentials at the beginning of the article. The steps are almost identical, just put "delta-u" instead of "du", or vice versa.
  2. There is already a rigorous proof given. In other words, there is already a "formal" derivation (using differentials), and a rigorous proof (using difference quotients, so unless this is really a new idea (and it isn't, see #1), why is it necessary?

Revolver 03:21, 20 Feb 2004 (UTC)

[edit] ???

This can also be seen as a barber shop analogy. For example, in the above example, u stands at one side while v takes the haircut.

Is this just silly vandalism, or am I really missing something??? Revolver 16:11, 30 September 2005 (UTC)
I couldn't understand this either and have removed it. Eric119 08:34, 16 October 2005 (UTC)

[edit] Leibniz and intuition

The argument of Leibniz can be adapted to provide better intuition than more common proofs.

Let ρ be an infinitesimal, and write g(x+ρ) as g(x)+g′(x)ρ+O(ρ2); likewise for f(x+ρ). Then (f·g)(x+ρ) will be

\begin{align} (f \cdot g)(x+\rho) &{}= f(x+\rho) \cdot g(x+\rho) \\ &{}= \left( f(x)+f'(x)\rho+O(\rho^2) \right) \cdot \left( g(x)+g'(x)\rho+O(\rho^2) \right) \\ &{}= f(x)g(x) + f'(x)g(x)\rho + f(x)g'(x)\rho + O(\rho^2) \end{align}

When we take the derivative we subtract off the f(x)g(x) term, divide by ρ, and discard any remaining terms containing ρ — which will be the O(ρ2) bit. In other words, the product rule is merely the distributive law followed by the discarding of higher order terms. This also quickly explains the result for a multiple product; any product term with two or more derivatives will be higher order, thus discarded.

The derivative here is, of course, a non-standard analysis version, of the form

f'(x) = \mathrm{standard}\left( \frac{f(x+\rho) - f(x)}{\rho} \right) ,

where ρ is an infinitesimal.

Sometimes this idea is conveyed with a picture of areas.

░░░░░░░░░░░░█
▒▒▒▒▒▒▒▒▒▒▒▒░
▒▒▒▒▒▒▒▒▒▒▒▒░
▒▒▒▒▒▒▒▒▒▒▒▒░
▒▒▒▒▒▒▒▒▒▒▒▒░
▒▒▒▒▒▒▒▒▒▒▒▒░
▒▒▒▒▒▒▒▒▒▒▒▒░

Here the top right corner corresponds to the higher order terms in ρ. --KSmrqT 04:26, 28 January 2007 (UTC)

[edit] multiplication signs, please

Formulas like

(fg)'=f'g+fg' \,
{d\over dx}(uv)=u{dv\over dx}+v{du\over dx}

are hard to decipher for the beginner.

Multiplication signs

(f\cdot g)'=f'\cdot g+f\cdot g' \,
{d\over dx}(u\cdot v)=u\cdot {dv\over dx}+v\cdot {du\over dx}

make it easier.

Bo Jacoby (talk) 21:43, 11 June 2008 (UTC).