Mathematics desk
< June 26	<< May \| June \| Jul >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 27

Product Rule Proof

I'm learning about the product rule, and I am trying to understand a common proof of the rule. I understand that one way to prove the product rule is as follows:

$h'(x)=\lim _{a\to 0}{\frac {h(x+a)-h(x)}{a}}=\lim _{a\to 0}{\frac {f(x+a)g(x+a)-f(x)g(x)}{a}}$

=\lim _{a\to 0}{\frac {f(x+a)g(x+a)-f(x)g(x+a)+f(x)g(x+a)-f(x)g(x)}{a}}

=\lim _{a\to 0}{\frac {[f(x+a)-f(x)]\cdot g(x+a)+f(x)\cdot [g(x+a)-g(x)]}{a}}

=\lim _{a\to 0}{\frac {f(x+a)-f(x)}{a}}\cdot \lim _{a\to 0}g(x+a)+\lim _{a\to 0}f(x)\cdot \lim _{a\to 0}{\frac {g(x+a)-g(x)}{a}}

=f'(x)g(x)+f(x)g'(x)

.

I'm wondering where $-f(x)g(x+a)+f(x)g(x+a)$ comes from in the numerator of the second step of the proof. I understand that both sets of terms cancel, but I am confused as to how one would know to multiply $f(x)$ and $g(x+a)$ and why this step is necessary.

Any help would be greatly appreciated.

Jtg920 (talk) 02:58, 27 June 2014 (UTC)[reply]

Well, it's needed to get to the next step. I'm not sure how the original concept behind the proof was discovered, but it most probably is due to Isaac Newton or Gottfried Leibniz, and it would be difficult to find their justification. — Arthur Rubin (talk) 04:03, 27 June 2014 (UTC)[reply]

Well, the concept is pretty obvious, isn't it? I mean, if you've seen it before. Just draw a square with base f(x) and height g(x), let x increase a tiny bit, and look at what happens to the area, ignoring the bit that's tiny in both dimensions.

Now, how to get from the concept to the algebra, that's just messing around until you get it right. --Trovatore (talk) 04:10, 27 June 2014 (UTC)[reply]

The article Product rule attributes it to Leibniz, and gives a reference, which appears no longer to be available online without a subscription or membership. — Arthur Rubin (talk) 04:12, 27 June 2014 (UTC)[reply]

The "trick" step is that

AB-ab=(A-a)B+a(B-b)

. To see this, consider a rectangle of sides A and B (we'll assume these are larger than a and b, respectively). Remove from that rectangle a smaller rectangle of sides a and b. The remaining L-shaped region can be cut into two rectangles of sides

(A-a),B

and

a,(B-b)

. Sławomir Biały (talk) 20:53, 27 June 2014 (UTC)[reply]

Adding and subtracting the same terms is a part of the "standard arsenal" of tricks that mathematicians use. If you're asking how on Earth anyone knew to do that (which was my reaction when I was taking calculus), the answer is probably just trial and error, along with experience about what has worked in similar situations. OldTimeNESter (talk) 20:11, 27 June 2014 (UTC)[reply]

The two middle terms

-f(x)g(x+a)+f(x)g(x+a)

are introduced for regrouping and factoring, a 'trick' which is somewhat similar to what you may recall from Algebra, in completing the square where you add a value so you must subtract the same value to keep the equation true. El duderino ^(abides) 07:31, 28 June 2014 (UTC)[reply]