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Product Rule

  1. Jun 30, 2007 #1
    We covered the product rule in maths last lecture, and as part of the derivation of it, we got this line:

    [tex]\frac{\partial y}{\partial x}=\frac{u \partial v}{\partial x} + \frac{v \partial u}{\partial x} + \frac{\partial u \partial v}{\partial x}[/tex]

    And were told that as [tex]x \rightarrow 0[/tex]

    [tex]\frac{\partial u \partial v}{\partial x}[/tex] just "goes away".

    Can anyone explain why this is so? :confused:

    (edit: quite pleased with my first attempt at tex :D )
  2. jcsd
  3. Jun 30, 2007 #2


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    I think you mean:
    [tex]\frac{dy}{dx}[/tex], instead of [tex]\frac{\partial y}{\partial x}[/tex]
    The [tex]\partial[/tex] 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:
    [tex]\frac{du dv}{dx} = \frac{du}{dx} dv = u'_x dv[/tex]

    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? :)
  4. Jun 30, 2007 #3
    darn, yea, swap all [tex]\partial[/tex] for [tex]\delta[/tex] etc... sorry. :blushing:

    Umm, not sure. After re-reading my notes, it says "Take limits as x -> 0".
    Last edited: Jul 1, 2007
  5. Jun 30, 2007 #4


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    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:

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

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

    [tex]\frac{f(x+h)g(x+h) - f(x)g(x+h)}{h} + \frac{g(x+h)f(x) - g(x)f(x)}{h}[/tex]

    And factoring out the common terms in each term:

    [tex]g(x+h)\frac{f(x+h) - f(x)}{h} + f(x)\frac{g(x+h)-g(x)}{h}[/tex]

    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, [tex]\left[f(x)g(x)\right]' = g(x)f'(x) + f(x)g'(x)[/tex]
  6. Jun 30, 2007 #5


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    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 [itex]\delta x[/itex] rather than [itex]\partial x[/itex] 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 [itex](\delta u \delta v)/\delta x[/itex]. Roughly speaking it is the fact that there are two [itex]\delta[/itex]'s in the numerator to only one in the denominator that makes it go to 0.
  7. Jul 1, 2007 #6
    I thought [tex]x \rightarrow 0[/tex] 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 [tex]\delta y / \delta x[/tex].

    The coordinates:


    [tex]Q=(x+ \delta x , (u+ \delta u)(v + \delta v))[/tex]


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

    [tex]\delta y=v \delta u + u \delta v + \delta u \delta v+ uv - uv[/tex]

    [tex]\frac{\delta y}{\delta x} = u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x} + \frac{\delta v \delta u}{\delta x}[/tex]

    Now, apparently, "Take limits as [tex]x \rightarrow 0[/tex]"

    [tex]\Rightarrow \frac{\delta y}{\delta x} = u \frac{\delta v}{\delta x} + v \frac{\delta u}{\delta x}[/tex]
  8. Jul 1, 2007 #7

    Gib Z

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    Could you perhaps explain it more rigorously? Because otherwise [tex]\int^b_a f(x) dx[/tex] would equal 0..for any bounds..or function...
  9. Jul 1, 2007 #8


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

    [tex]\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[/tex]

    Now, take the limit as [tex]\delta x \rightarrow 0[/tex], we have:

    [tex]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)[/tex], since, u and v are differentiable, [tex]\lim_{\delta x \rightarrow 0} \frac{\delta u}{\delta x} = u'(x)[/tex], [tex]\lim_{\delta x \rightarrow 0} \frac{\delta v}{\delta x} = v'(x)[/tex], both are finite. So we have:

    [tex]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[/tex]

    Since v is continuous, so, as [tex]\delta x \rightarrow 0[/tex], we also have: [tex]\delta v \rightarrow 0[/tex], so:

    [tex]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)[/tex]

    Well, that's how I understand it.

    I did mistype in the previous post, all dx, dy, du, or dv should be [tex]\delta x[/tex], [tex]\delta y[/tex], [tex]\delta u[/tex], and [tex]\delta v[/tex]. Sorry for the confusion. :blushing: :redface: :)
    Last edited: Jul 1, 2007
  10. Jul 2, 2007 #9

    Gib Z

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    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 [tex]\int^b_a f(x) dx[/tex] 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 :(
  11. Jul 2, 2007 #10

    matt grime

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    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 [itex]\delta y /\delta x[/itex] (note you use itex and not tex for inline tags) it is the slope of a chord, as you let [itex]delta x[/itex] 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 [itex]\delta x[/itex] tends to 0, not x. Then it is correct. Remember that [itex]\delta u[/itex] is approximately [itex]\delta x .du/dx[/itex], similarly for v, thus the cross term is approximately

    [tex] \delta x \frac{du}{dx}\frac{dv}{dx}[/tex]

    and does indeed tend to zero as [itex]\delta x[/itex] tends to 0
  12. Jul 2, 2007 #11


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    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?

    [tex]\lim_{n \rightarrow \infty} n \times \frac{1}{n} = ?[/tex]

    Would it be 0? :smile:
    Last edited: Jul 2, 2007
  13. Jul 2, 2007 #12

    Gib Z

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