# Differentiating Products

1. Jul 29, 2008

### Dumbledore

Hello, in coming up with a general form for finding the derivative of a product there is one step I do not understand.

if we let u, v and y be functions of x then:

dy/dx = u(dy/dx) + v(du/dx) + du (dv/dx)

From there we take the lim as dx -> 0 of each term and the textbook I am reading says:

"Now let dx -> 0. Then du -> 0"

I simply do not understand why it follows that du -> 0 because dx -> 0.

With this assumption however we are able to find the simplified general form:

dy/dx = u(dv/dx) + v(du/dx)

I applogize for not using the fancy math notation symbols. I have never used them before so I don't really know what half of them mean. I hope you guys can understand my question better than I can...

2. Jul 29, 2008

### tiny-tim

Hi Dumbledore!

(have a curly-wurly delta: δ)
You mean dy/dx = u(dv/dx) + v(du/dx) + du (dv/dx).

I think it means that if you move from x to x + δx, so that u v and y move to u + δu, v + δv, and y + δy,

then obviously (y + δy) - y = (u + δu)(v + δv) - uv = uδv + vδu + δuδv.

Now divide by δx: (uδv + vδu + δuδv)/δx

and let δx tend to zero (so all the other δs tend to zero also), and you get:

u dv/dx + v du/dx + δu dv/dx,

= u dv/dx + v du/dx + 0.

(or, of course, you can write it u dv/dx + v du/dx + δv du/dx.)

That's the δu which they're talking about.

3. Jul 29, 2008

### Dumbledore

I think the part that I was missing is that if δx approaches 0 then all the other δs approach zero. Now it makes perfect sense. Thanks tiny tim!