# Product Rule

1. Jun 30, 2007

### joshd

We covered the product rule in maths last lecture, and as part of the derivation of it, we got this line:

$$\frac{\partial y}{\partial x}=\frac{u \partial v}{\partial x} + \frac{v \partial u}{\partial x} + \frac{\partial u \partial v}{\partial x}$$

And were told that as $$x \rightarrow 0$$

$$\frac{\partial u \partial v}{\partial x}$$ just "goes away".

Can anyone explain why this is so?

(edit: quite pleased with my first attempt at tex :D )

2. Jun 30, 2007

### VietDao29

I think you mean:
$$\frac{dy}{dx}$$, instead of $$\frac{\partial y}{\partial x}$$
The $$\partial$$ one is used to denote partial derivative, the one you are working with in muti-variable calculus.

Are you sure it's not dx ~~> 0?

Ok, you can think like this:
$$\frac{du dv}{dx} = \frac{du}{dx} dv = u'_x dv$$

du/dx is the derivative of u, with respect to x, is a finite number, multiply with dv, a very very small number (since dv also tends to 0, as dx tends to 0.). So the whole thing should be 0. So, it just "go away". :)

Can you get it? :)

3. Jun 30, 2007

### joshd

darn, yea, swap all $$\partial$$ for $$\delta$$ etc... sorry.

Umm, not sure. After re-reading my notes, it says "Take limits as x -> 0".

Last edited: Jul 1, 2007
4. Jun 30, 2007

### Mute

I'm not sure how your professor derived the product rule, so I don't know how that cross-term comes about in your derivation there. Starting from the definition of the derivative and using the "add zero" trick, the proof is easy:

$$\left[f(x)g(x)\right]' = \lim_{h\rightarrow 0}~\frac{f(x+h)g(x + h) - f(x)g(x)}{h}$$

Then introduce the terms $$g(x+h)f(x) - f(x)g(x+h)$$ in the numerator to get (limit notation left out):

$$\frac{f(x+h)g(x+h) - f(x)g(x+h)}{h} + \frac{g(x+h)f(x) - g(x)f(x)}{h}$$

And factoring out the common terms in each term:

$$g(x+h)\frac{f(x+h) - f(x)}{h} + f(x)\frac{g(x+h)-g(x)}{h}$$

As h -> 0, the fractions tend to derivatives and g(x+h) tends to g(x) (since the limit of a product is the product of the limits)

So, $$\left[f(x)g(x)\right]' = g(x)f'(x) + f(x)g'(x)$$

5. Jun 30, 2007

### HallsofIvy

I would consider that extremely bad notation! Neither "x" nor "dx" "goes to 0". Certainly x doesn't and if you are using "dx" as a differential, then it doesn't "go to 0"- it is and remains an infinitesmal. With $\delta x$ rather than $\partial x$ or dx it makes a little more sense but you didn't say how y is a function of u and v (nor what u and v mean) so I can't tell how you would get $(\delta u \delta v)/\delta x$. Roughly speaking it is the fact that there are two $\delta$'s in the numerator to only one in the denominator that makes it go to 0.

6. Jul 1, 2007

### joshd

I thought $$x \rightarrow 0$$ means "x tends to 0", which means x is infinitly small, it doesn't mean that x IS 0?

I know I miseed alot of stuff out, hence why I just said the line was just part of it...

We started with a curve, equation y=u.v, with two points, P and Q. The gradient between them is $$\delta y / \delta x$$.

The coordinates:

$$P=(x,u.v)$$

$$Q=(x+ \delta x , (u+ \delta u)(v + \delta v))$$

So:

$$\delta y = (u+ \delta u)(v+ \delta v)-(uv)$$, where $$(u+ \delta u)(v+ \delta v)$$ is the value of the curve at Q and $$(uv)$$ is the vale of the curve at P.

$$\delta y=v \delta u + u \delta v + \delta u \delta v+ uv - uv$$

$$\frac{\delta y}{\delta x} = u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x} + \frac{\delta v \delta u}{\delta x}$$

Now, apparently, "Take limits as $$x \rightarrow 0$$"

$$\Rightarrow \frac{\delta y}{\delta x} = u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x}$$

7. Jul 1, 2007

### Gib Z

Could you perhaps explain it more rigorously? Because otherwise $$\int^b_a f(x) dx$$ would equal 0..for any bounds..or function...

8. Jul 1, 2007

### VietDao29

Well, let y = uv, where u, and v are functions of x, and everywhere differentiable.
So: $$\delta y = (u + \delta u) (v + \delta v) - uv = u \delta v + v \delta u + \delta u \delta v$$
Divide both sides by $$\delta x$$, we obtain:

$$\frac{\delta y}{\delta x} = u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x} + \frac{\delta u \delta v}{\delta x} = u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x} + \frac{\delta u}{\delta x} \delta v$$

Now, take the limit as $$\delta x \rightarrow 0$$, we have:

$$y'(x) = \lim_{\delta x \rightarrow 0} \frac{\delta y}{\delta x} = \lim_{\delta x \rightarrow 0} \left( u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x} + \frac{\delta u}{\delta x} \delta v \right)$$, since, u and v are differentiable, $$\lim_{\delta x \rightarrow 0} \frac{\delta u}{\delta x} = u'(x)$$, $$\lim_{\delta x \rightarrow 0} \frac{\delta v}{\delta x} = v'(x)$$, both are finite. So we have:

$$y'(x) = \lim_{\delta x \rightarrow 0} \frac{\delta y}{\delta x} = \lim_{\delta x \rightarrow 0} \left( u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x} + \frac{\delta u}{\delta x} \delta v \right) = u(x)v'(x) + u'(x)v(x) + u'(x) \lim_{\delta x \rightarrow 0} \delta v$$

Since v is continuous, so, as $$\delta x \rightarrow 0$$, we also have: $$\delta v \rightarrow 0$$, so:

$$y'(x) = \lim_{\delta x \rightarrow 0} \frac{\delta y}{\delta x} = ... = u(x)v'(x) + u'(x)v(x) + u'(x) 0 = u(x)v'(x) + u'(x)v(x)$$

Well, that's how I understand it.

I did mistype in the previous post, all dx, dy, du, or dv should be $$\delta x$$, $$\delta y$$, $$\delta u$$, and $$\delta v$$. Sorry for the confusion. :)

Last edited: Jul 1, 2007
9. Jul 2, 2007

### Gib Z

Ok I seem to be able to understand that, your reasoning is that a finite value ( u'(x)) multiplied by an infinitesimal that is approaching zero, must also be zero. However, wouldn't that mean that when we take the integral $$\int^b_a f(x) dx$$ We are summing up an infinite number of slices between b and a, each valued f(x) dx, as dx approaches zero? By the previous logic every integral is zero :(

10. Jul 2, 2007

### matt grime

it does

it doesn't. That phrase is meaningless.

correct - one cannot just insert x=0 into an object whose limit you're taking - that negates the whole point of taking limits in the first place

That is not the derivative. For a given $\delta y /\delta x$ (note you use itex and not tex for inline tags) it is the slope of a chord, as you let $delta x$ tend to zero it will (if the function is differentiable) converge to the gradient.

Two things - that is an horrendous abuse of an equals sign. A problem that infests lower level mathematics teaching. It is true _approximately_.
Secondly, that should read as $\delta x$ tends to 0, not x. Then it is correct. Remember that $\delta u$ is approximately $\delta x .du/dx$, similarly for v, thus the cross term is approximately

$$\delta x \frac{du}{dx}\frac{dv}{dx}$$

and does indeed tend to zero as $\delta x$ tends to 0

11. Jul 2, 2007

### VietDao29

Nope. You should note that it's the sum of an infinite numbers of strips.

Say, if you sum (1/n), n times, as n increases without bound, what would you get?

$$\lim_{n \rightarrow \infty} n \times \frac{1}{n} = ?$$

Would it be 0?

Last edited: Jul 2, 2007
12. Jul 2, 2007

### Gib Z

I'm sorry, i do have lapses of idiocy :) Even wrote up the keyword and didn't realise it lol, thanks !