Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why do we learn differentials?

  1. Nov 29, 2003 #1


    User Avatar
    Science Advisor

    One application of derivatives from first year calculus is something called differentials. The intent is to find the change of something based on the derivative of a function and some sort of varialbe like time or a distance or something.
    Let's say you have this formula:
    y = x^2
    now here is the derivative:
    dy/dx = 2x
    now if you bring the dx over, it looks like this
    dy = 2x dx

    In math class, these are meant to find changes in things. Let's say you wanted to find the change in y when x changes from 5 to 10. you would just fill in the equation like this:
    dy = 2(5)(5)
    dy = 50

    the dy is your change in y. the first 5 is your original x value. the second 5 is your change in x.

    The differential said the change is 50. Now lets see what the original equation says the difference is:
    final - original
    = x^2 - x^2
    = 10^2 - 5^2
    = 100 - 25
    = 75

    The two different equations give VERY different answers. They're not even close. Knowing this, why do we still learn these?
  2. jcsd
  3. Nov 29, 2003 #2


    User Avatar
    Science Advisor
    Gold Member

    dy/dx is the instantaneous rate of change of y with respect to x. Since y(x) = x2 is nonlinear, you can't expect dy/dx to equal Δy/Δx
  4. Nov 29, 2003 #3


    User Avatar
    Science Advisor

    Exactly. If they don't work then why the hell do we learn them?
  5. Nov 29, 2003 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because differentials work well then the quantities involved are small.

    Let's stick to the [itex]f(x)=x^2[/itex] example, but with a smaller differential... how about [itex]x=10[/itex] and [itex]\delta x=1[/itex].

    In this case, we have [itex]f(11)=121[/itex] and the differential approximation gives [itex]f(11) \approx f(10) + 1 * f'(10) = 120[/itex] which is pretty darn close.

    The relevant theorem is:

    f(x + \delta x) = f(x) + f'(x) \delta x + \varepsilon (\delta x) \delta x \
    \mathrm{where} \lim_{\delta x \rightarrow 0} \varepsilon(\delta x) = 0

    In other words, the error term in the approximation of [itex]f(x+\delta x)[/itex] shrinks "quickly" as [itex]\delta x[/itex] approaches 0. In fact, if [itex]f(x)[/itex] is twice differentiable, you can prove that there exists a constant [itex]c[/itex] such that [itex]|\varepsilon (\delta x)| < c |\delta x|[/itex], so the error term is quadratic in [itex]\delta x[/itex]. (This is the Taylor remainder theorem; a differential approximation is just a first degree Taylor polynomial!)

    edit: finally got the LaTeX right. :smile:
    Last edited: Nov 29, 2003
  6. Nov 29, 2003 #5


    User Avatar
    Science Advisor

    oh ok, that makes sense.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook