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Homework Help: Question about derivatives and differentials

  1. Apr 8, 2014 #1
    I have an easy question which I've been thinking about for a while..

    Lets say I want to take the derivative of a function y = f(x) with respect to x, we would get.

    dy/dx = f'(x).

    In the couple of books I've skimmed through, they all say that dy/dx is not a ratio but the notation that implied taking the derivative of y with respect to x.

    If dy/dx is not a ratio then how come the differential of y is equal to f'(x)dx? It almost seems as they are multiplying both sides by dx. This can't be mathematically correct, can it? I would like to know mathematically how dy is equal to f'(x)dx.
  2. jcsd
  3. Apr 8, 2014 #2


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  4. Apr 10, 2014 #3


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    Less formally, yes, the derivative, dy/dx, is NOT a fraction but it is the limit of a fraction and as a result can often be treated as a fraction (go back before the limit, use the fraction property, and then take the limit again). In order to make that explicit, we define differentials by "dy = f'(x)dx. Strictly speaking, neither "dy" nor "dx" is defined separately- they must appear together.

    (In differential geometry where we have vector functions or function values of other kinds of things than numbers "differentials" are given a slightly different (but equivalent) and more precise definition.)
  5. Apr 10, 2014 #4
    For practical purposes, you can almost always treat dy/dx as a ratio.
    When you learn the chain rule, it says "dy/dx = (dy/du)(du/dx)" -- and you'll say, well, of course. Later on, when you come to partial derivatives, you have the seemingly strange result that the product of two partials that look like you could just "cancel" one differential will turn negative. So be a bit wary. But in the first calculus class, when you'll be doing "related rate" problems, you can confidently write things like,
    dh/dt = (dV/dt)/(dV/dh), as if derivatives were ordinary fractions.
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