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Splitting a Derivative

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  1. Oct 29, 2015 #1
    After doing a couple courses in physics as well as calculus and differential equations, I was starting to wonder about splitting a derivate, such as ## \frac{dy}{dx} ##, into seperate pieces ##dy## and ##dx##. I know we've never done it in calculus or differential equations because it isn't technically a fraction, its one thing. But if that's the case, why does it work when doing it in more applied courses such as physics? Are there certain rules on when this kind of thing works and does not?
     
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  3. Oct 30, 2015 #2

    andrewkirk

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    The general rule is that it is permissible to do it when you can prove that there is a way to arrive at the same result without doing it.

    For example, it is fairly easy to prove that ##\frac{dy}{dt}/\frac{dx}{dt}=\frac{dy}{dx}## provided all three derivatives exist and ##\frac{dx}{dt}## is nonzero. The proof just uses the definition of derivative in terms of limits, together with properties of limits of products and quotients.

    After a while you get to know what these valid uses are. If in doubt, don't just assume you are allowed to do it.
     
  4. Nov 1, 2015 #3

    HallsofIvy

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    There are a number of different ways of handling this. The simplest is to define "dx" as a purely symbolic "differential", then define dy= f'(x) dx. Although you are correct that the derivative is NOT fraction it is a limit of fractions. By going back before the limit, using fractions properties then taking the limit, one can show that the derivative can be treated like a fraction. The "differential" notation above formalizes that.
     
  5. Nov 4, 2015 #4
    dx just means change in x. dy/dx is change in y relative to a change in x. with integrals it is known as the integrand and it refers to the variable that is being integrated. Your supposed to understand that in calc 2 and 3 but seeing it in an application is not so easy. Its like an infinitely small change moreover and heck yea you can split that ish up screw the rules.
     
  6. Nov 4, 2015 #5

    Mark44

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    No, that would be ##\Delta x##. dx is the differential of x.
    dy/dx is one form of notation for the derivative of y with respect to x. It is defined as a limit.
    $$\frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$
    The change in y relative to a change in x is ##\frac{\Delta y}{\Delta x}##.
    No, dx is not the integrand. In an indefinite integral such as ##\int 3x^2 + 2x~dx##, the integrand is ##3x^2 + 2x##. The dx part has different meanings for different kinds of integrals (e.g., Riemann integrals, Rieman-Stiltjes integrals, Lebesgue integrals).
    From wikipedia (https://en.wikipedia.org/wiki/Integral):
     
  7. Nov 4, 2015 #6
    I don't understand I thought the dx refers to the variable that is being integrated, thank you for correcting me on the integrand. And yes I understand the integral as a sum such that dx would be finite and Tri-x. Moreover what's the big deal with dx not meaning change in x?
     
  8. Nov 4, 2015 #7
    if you a function F(x) it will change relative to x. The change in F(x), dF/dx, is another function that changes relative to x. I don't see the need to say its the differential of x because its continuous. Isn't it the same thing anyways, and please tell me its not because every math teacher gets pissed off when you overlook dx. Please Mark44 help me understand, you can explain the different meanings I know you can.
     
  9. Nov 4, 2015 #8

    Mark44

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    Yes, dx conveys that information in a Riemann integral. It has a different meaning in the other integrals I mentioned.
    What is Tri-x?
    dx is an infinitesimal, while ##\Delta x## means a small, but finite, change in x. In practice, dx and ##\Delta x## are treated as synonomous, but the same is not true for dy and ##\Delta y##.

    In the picture below, the blue curve is supposed to be the graph of y = f(x). At the lower left corner of the triangle is the point ##(x_0, y_0)##, where ##y_0 = f(x_0)##. Partway up the vertical line is the distance dy, where dy = f'(x0) * dx. Farther up, the vertical line intersects the curve at ##(x_1, y_1)##, where ##y_1 = f(x_1)##, and ##x_1 = x_0 + dx = x_0 + \Delta x##.

    We can estimate ##y_1## using the tangent line, as ##y_1 \approx y_0 + dy = y_0 + f'(x_0) dx##. Notice that because the curve is concave up, our estimate for ##y_1## will be smaller than the true value of ##y_1##, which would be ##y_1 = y_0 + \Delta y##.
    Snapshot.jpg
     
  10. Nov 4, 2015 #9

    Mark44

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    There's a verb missing here. What are you trying to say?
    No, this isn't right. The change in F is ##\Delta F##. DF/dx is the derivative of F with respect to x.
    Let's say we're talking about two points (x1, F(x1)) and (x2, F(x2)), where x2 - x1 = ##\Delta x##.

    The change in F is ##\Delta F## (or ##\Delta y## if y = F(x)), the change in the y values of function F. IOW, this is F(x2) - F(x1). This is only approximately equal to dF, and only when x2 - x1 is "small." The picture I drew in my other post shows the difference between dy and ##\Delta y##.
     
  11. Nov 4, 2015 #10
    delta x

    You know me too well, I think I get it and I agree but I just think with very small delta x delta y approaches dy. Thank you.
     
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