# Looking for a few good proofs

1. Mar 7, 2004

### Matt Jacques

I understand how to use such things as product rules, quotient rules, parts by integration, but it bothers me I dont really have a deeper understanding of it.

My book offers rather rigorous proofs, they are all pretty much: assume this to be this and let this be that so it must equal this. Hmm. Ya.

Does anyone know of any good sites that proves them in understandable and no omitted steps?

2. Mar 7, 2004

### matt grime

I'm sorry, I don't understand. How can your book simultaneously be too rigorous, yet omit steps? It may perhaps omit obvious calculations - it is up to you to put them back in if you can't see them.

If you want motivation as to why the product rule is true, is this either too rigorous, or omitting too many steps?

f is diffible at x if

f(x+h) = f(x) +hf'(x) +h*o_1(h)

where o(h) is something that tends ot zero as h tends to zero. f' is defined to be the derivative

then
f(x+h)g(x+h)=(f(x)+hf'(x)+h*o_1(h))(g(x)+hg'(x)+h*o_2(h))

multiply out:

f(x)g(x)+ h(f(x)g'(x)+f'(x)g(x)+ h*o_3(x))

where o_3(x) = hg'(x)f'(x)+hf'(x)o_2(x)+hg'(x)o_1(x)+ho_1(x)o_2(x)+f(x)o_2(x)+g(x)o_1(x)

which is a function that tends to zero as h tends to zero.

Hence the derivative of f(x)g(x) is f'(x)g(x)+f(x)g'(x)

similar analysis allows you to prove the chain rule (messy) and the quotient rule, which actually just follows from the previous two. You should prove the chain rule - a liberal disrespect for the quantities you manipulate is to be encouraged.

Last edited: Mar 7, 2004