# Proof of the product rule

• Dec 31st 2012, 08:19 AM
kinhew93
Proof of the product rule
I've just been looking at a proof of the product rule and there is one part which I can't get my head round:

Part of the proof states that when δx tends to 0, so does δu, δv and δy. I'm fine with this.

What I don't get is that when δx tends to 0, (δuδv)/δx also tends to 0. I don't see whats going on here (I never really learnt about limits formally) so if possible could somebody please try to explain why this is?

Thanks :]
• Dec 31st 2012, 08:51 AM
emakarov
Re: Proof of the product rule
Basically, δu/δx tends to du/dx and δv tends to 0, so the product of those expressions tends to the product of limits, i.e., 0. You can see this explained in the proof in Wikipedia. Search for the paragraph starting with "The third term" in that section.
• Dec 31st 2012, 01:04 PM
Deveno
Re: Proof of the product rule
roughly speaking, it's because when Δu and Δv are small, ΔuΔv is "even smaller".

Δu/Δx is almost the same as u'(x0) (we can make it as close as we like by making Δx small). and for any given x0, this is some number, call it a.

then ΔuΔv/Δx = (Δu/Δx)Δv is very close to aΔv, which is very close to a*0 = 0.

let's take a real function, and see how this works:

suppose y = f(x) = x2sin(x), so:

u(x) = x2
v(x) = sin(x)

first we'll look at Δy/Δx for Δx = 1/1000, and x0 = 1.

Δy/Δx = [f(1.001) - f(1)]/0.001 = [(1.001)2(sin(1.001)) - sin(1)]/0.001 ~ [(1.002001)(0.84255044437568436165660998445205) - (0.8414709848078965066525023216303)]/(0.001)

~ 0.0027654030069835994117825394006425/0.001 ~ 2.7654030069835994117825394006425

now Δy = y(x0+Δx) - y(x0) = u(x0 + Δx)v(x0 + Δx) - u(x0)v(x0)

= u(x0 + Δx)v(x0 + Δx) - u(x0)v(x0 + Δx) - u(x0)v(x0) + u(x0)v(x0 + Δx)

= (u(x0 + Δx) - u(x0))(v(x0 + Δx)) + (u(x0))(v(x0 + Δx) - v(x0))

= (Δu)(v(x0 + Δx)) + (u(x0))(Δv)

= (Δu)(v(x0 + Δx)) - (Δu)(v(x0)) + (u(x0))(Δv) + (Δu)(v(x0))

= (u(x0))(Δv) + (Δu)(v(x0)) + ΔuΔv

thus:

Δy/Δx = [(u(x0))(Δv)]/Δx + [(Δu)(v(x0)]/Δx + (ΔuΔv)/Δx

let's see approximately what each of these 3 terms are:

first we'll see what Δu and Δv are (approximately).

Δu = [(1.001)2 - 12] = 0.002001

Δv = [sin(1.001) - sin(1)] ~ (0.84201086628825692390267737345894) - (0.8414709848078965066525023216303) ~ 0.0053988148036041725017505182864499

so [(u(x0))(Δv)]/Δx ~ 0.00053988148036041725017505182864499/(0.001) = 0.53988148036041725017505182864499 <--this is significant (nowhere near 0, compared to Δx)

and [(Δu)(v(x0)]/Δx = (0.002001)(sin(1))/(0.001) ~ (2.001)(0.8414709848078965066525023216303) ~ 1.6837834406006009098116571455822 <--this is also significant (right? RIGHT??)

finally: (ΔuΔv)/Δx ~ (0.002001)(0.0053988148036041725017505182864499)/(0.001) ~ 0.0010803028422011949176002787091186 <---about the same size as Δx (Δx can be thought of as "how accurately we can measure our input". numbers smaller than Δx are insignificant).

in the language of infinitesimals, dudv is infinitesimally small compared to dx, which is infinitesimally small compared to x, so we can disregard it (believe it or not, before modern analysis (cauchy, weierstrass, dedekind, et alia) this was the way that derivatives were actually justified...it was believed by many that this was "bad mathematics" and it wasn't until the mid 20-th century that it became "OK" to think this way again).
• Dec 31st 2012, 01:13 PM
Soroban
Re: Proof of the product rule
Hello, kinhew93!

I can give you the traditional proof of the Product Rule
. . using the definition: . $f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}$
I've taught it a few times and can still recreate it from memory.

We are given: . $P(x) \;=\;f(x)\!\cdot\!g(x)$

$f(x\!+\!h) - f(x) \:=\:f(x\!+\!h)\!\cdot\!g(x\!+\!h) - f(x)\!\cdot\!g(x)$

Subtract and add $f(x\!+\!h)\!\cdot\!g(x)$

$f(x\!+\!h)-f(x) \:=\: f(x\!+\!h)\cdot g(x\!+\!h) {\color{blue}- f(x\!+\!h)\!\cdot\!g(x) + f(x\!+\!h)\!\cdot\!g(x)} - f(x)\cdot g(x)$

n . . . . . . . . $=\;f(x+h)\bigg[g(x+h) - g(x)\bigg] + g(x)\bigg[f(x+h) - f(x)\bigg]$

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

$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
. . . $=\; \underbrace{\lim_{h\to0}f(x+h)}_{f(x)}\cdot \underbrace{\lim_{h\to0}\frac{g(x+h) - g(x)}{h}}_{g'(x)}\;+\;g(x)\cdot\underbrace{\lim_{h \to0} \frac{f(x+h)-g(x)}{h}}_{f'(x)}$

Therefore: . $P'(x) \;=\;f(x)\!\cdot\!g'(x) + f'(x)\!\cdot\!g(x)$
• Dec 31st 2012, 01:20 PM
emakarov
Re: Proof of the product rule
I have a followup problem: Name MHF members who either have already prepared their New Year party or have nowhere to go, and therefore have too much free time on their hands. (Sadsmile)