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Leibniz Notation

  1. Dec 14, 2015 #1
    The entire first semester of my Calculus class we used Lagrange's notation, f'(x), f''(x), etc.. So at the beginning of second semester the teacher kinda casually switched over to Leibniz notation, dy/dx, which left all of the class dazed.

    I understood it pretty well until she did a simple problem, dy/dx=2x/y. She asked us to find the anti-derivative of it so I cross multiplied. (dx)(2x)=(dy)(y). The way I understand it is that dx or dy is an infinitesimal change in x or y. So an infinitesimal change in x times 2x would approach 0 and an infinitesimal change in y times y would approach 0 too. Then I can't take the anti-derivative. I need a better explanation of this because obviously I'm doing something wrong. Any help is welcomed thanks.
  2. jcsd
  3. Dec 14, 2015 #2


    Staff: Mentor

    You separated the original equation, dy/dx = 2x/y, to get 2xdx = y dy. The next step is to integrate the left side with respect to x and the right side with respect to y. IMO, thinking about dx and dy as infinitesimals is not helpful. Instead, think of them as indicating which is the variable of integration.
  4. Dec 14, 2015 #3


    User Avatar
    Gold Member

    To clarify:
    Perhaps consider the derivative. This is, for example, the instantaneous change in speed (y) at a certain point in space (x). We can denote the derivative as being dy/dx...

    Read literally, as you did, this means 'infinitely small change in y divided by some infinitely small change in x'. You'll get some ratio whose value is the instanaeous slope at that point x. BUT, the same way for the integral, you could say this is 0/0 but that is not really helpful.. it doesn't tell you the instantaneous rate of change.

    Hope that clarifies.
  5. Dec 14, 2015 #4
    Lagrange and Leibniz notation mean the same thing. I personally use Lagrange when just doing basic derivatives, but Leibniz notation is more useful in my opinion when it comes to implicit differentiation, multi-variable calculus (especially with partial derivatives), and differential equations. I personally prefer to write in Leibniz notation over LaGrange. It's easier to keep track of the variables. And I think that Leibniz notation is more effective when teaching U-substitution to students. My professor used LaGrange to teach U-substitution which made it confusing until I went home and practiced it with Leibniz notation.

    Here's what I mean:
    Consider the following integral:
    \int_{a}^{b} \frac{cos(ln(x)}{x} dx
    Let u = ln(X) such that
    du/dx (The derivative of u with respect to x) = (1/x)
    du = 1/x dx
    See how easier it is to keep track of variables with Leibniz notaiton. dy/dx is just saying the derivative of y with respect to x.

    Likewise, partial differential can be written in terms of both Leibniz notation and LaGrange notation.
    \frac{\partial f}{\partial x} = f_x
    Where Leibniz notation is on the left and LaGrange notation is on the right.

    My point for my long reply is that don't let notation scare you. Just use the notation that you feel most comfortable with. And that you might find LaGrange more useful for some concepts and Leibniz for others.
  6. Dec 14, 2015 #5
    Thank you all for the help.
    I do agree I think Leibniz notation would be much more organized, but what do I need to see when I see du = 1/x dx. What does that mean. Is that a useful form or does it need to be altered.
  7. Dec 14, 2015 #6
    When evaluating an integral using substitution you need to change the variable to u and integrate with respect to du instead of dx.

    \int_{a}^{b} \frac{cos(ln(x)}{x} dx
    If we let u = ln(x) we can take the diverative of u (our function) with respect to x which is written as (du/dx)
    \frac{du}{dx} = \frac{1}{x}
    If we move dx to the other side, we get a value for du
    du = \frac{1}{x} dx
    We can replace ln(x) in the intergal with u since we defined it as such. And we can replace 1/x dx with du since it is defined as such.
    Therefore, our intergal becomes
    \int_{a}^{b} cos(u) du
    Now that we have our intergal in terms of u, we can easily integrate with with respect to u.
    \int_{a}^{b} cos(u) du = sin(u)
    We replace u back with ln(x), we get the final answer to be sin(lnx)

    To answer your question, du is not always equal to 1/x dx, it depends on how you define u. But writing du/dx and then moving dx to the other side, allows you to define du in terms of dx which you are trying to eliminate so you can integrate with respect to u. You replace some f(x) dx with du so you can make the integral simpler and then integrate with respect to u. In essence, you're undoing the chain rule.

    Consider this example
    \int_{a}^{b} 2x e^{x^2} dx.
    You don't know how to integrate this without trying to simplify the integral.
    So we can let u = x^2 so that the derivative of u with respect to x (written as du/dx) as:
    \frac{du}{dx} = 2x
    Moving dx to the other side, we get
    du = 2x dx
    We see that we can replace 2x dx in our integral with du.
    Therefore our integral becomes:
    \int_{a}^{b} e^{u} du.
    Solving our integral we get:
    \int_{a}^{b} e^{u} du. = e^{u} = e^{x^2}
    so our final answer is e^{x^2}
    So, Leibniz notation is good for keep track of variables with u-substitution.
  8. Dec 15, 2015 #7
    Ok, thank you that clears it up very well :)
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