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Differentiation: the prime notation

  1. Jun 29, 2013 #1
    Hi,

    I have been using, for the most part, the prime notation when I want to indicate differentiation. As off recently, I have gained more insight into Leibniz's notation. This triggered the following question: how does the prime notation indicate what we are differentiating with respect to? I immediately thought that the answer to my question is: whatever number goes in between the brackets e.g. f'(u) is the derivative of f with respect to u.

    Is that accurate?

    Thanks!
     
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  3. Jun 29, 2013 #2

    micromass

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    The prime notation doesn't specify what we're differentiating to. This is the main disadvantage of the notation. It should always be made clear what the variable is.
     
  4. Jun 29, 2013 #3
    Thank you for the prompt reply!

    One of the reasons I asked this was to try and make sure I was translating the chain rule from prime to Leibniz notation correctly.

    If the term in brackets indicated what we are differentiating to, it would be pretty easy. If that is not the case, I have a hard time proving to myself that:

    Given y = f(u) and u = g(x)

    (f (g (x)) )' = dy/dx and f'(g (x)) = dy/du

    Could you help me with that please?
     
  5. Jun 29, 2013 #4

    lurflurf

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    So in each notation the chain rule is written

    $$(\mathop{f}(\mathop{g}(x)))^\prime={\mathop{f}} ^\prime (\mathop{g}(x)) \mathop{g} {}^\prime (x) $$

    $$\dfrac{df}{dx}=\dfrac{df}{dg} \dfrac{dg}{dx} $$

    A slight abuse of notation is common in which we do not distinguish between functions and variables
     
  6. Jun 30, 2013 #5
    (Disclaimer: My answer is pretty similar to what lurflurf's said.) It's important to note that here [itex]u = u(x)[/itex]. If it's confusing, think of [itex]u[/itex] as a function, not a variable. It's like you're passing the whole function [itex]u(x)[/itex] in as an input to the function [itex]f[/itex]. Or think of [itex]u[/itex] as just being a shorter way of writing [itex]g(x)[/itex].

    So, the chain rule is just:

    [itex]y = f(g(x)).[/itex]​

    and

    [itex](f(g(x)))' = f'(g(x)) \cdot g'(x).[/itex]​


    When we rewrite this in [itex]\frac{dy}{dx}[/itex] notation, we put the "independent variable"/"function input" in the bottom, and the "dependent variable"/"function" in the top. So for [itex]f'(g(x))[/itex], [itex]f[/itex] goes in the top, which is just [itex]y[/itex], and [itex]g(x)[/itex] goes in the bottom, which is just [itex]u[/itex]. For [itex]g'(x)[/itex], [itex]g[/itex] goes in the top, which is just [itex]u[/itex], and [itex]x[/itex] goes in the bottom. So we have

    [itex]\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}[/itex]​
     
  7. Jul 1, 2013 #6

    Stephen Tashi

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    The only accurate way to interpret that notation in words is to say that [itex] f'(.) [/itex] means the derivative of f with respect to the argument of the function. Don't get hypnotized by particular letters. For example, to define a function f(u) = 2u + 5 is the same thing as to define it as f(x) = 2x + 5. Specific letters don't become crucial until you begin discussing a problem where 'u' and 'x' represent different things.

    For example, the following problem is easy:

    Given:
    f(u) = 2u + 5
    h(x) = x^2 + x
    Find the derivative of f(h(x)).

    The following problem confuses some students
    Given
    f(x) = 2x + 5
    h(x) = x^2 + x
    Find the derivative of f(h(x))

    The two problems above are the same problem.

    This following can be considered a "trick" problem
    Given
    f(u) = 2u + 5
    h(x) = x^2 + x
    g(x) = f(h(x))

    Find g'(u).

    This problem , in my opnion, is tricky since it might occur in a larger context where 'x' and 'u' have been given some definite and different interpretations (e.g. that x is voltage and u is current) and also given some equation relating 'x' and 'u'. In absence of any context, 'x' and 'u' are merely used as "place holders" in defining the functions.

    Defining a function in terse manner, such as "f(u) = 2u + 5" is an abbreviation for more lengthy statement. The "place holder" interpretation of "f(u) = 2u + 5" is "For each number u, the function f maps u to 2u + 5". The "variable" u only has meaning "inside" that statement. (i.e. The "scope" of the variable u is that statement alone.

    If you are familiar with computer programming, you know that variable names like "u" have a certain "scope" in complicated programs. The same name might be used in different sections of the program and there is no relation enforced between the different uses of same name. The same is true of written mathematics, However mathematical writing is more sloppy and you often must struggle to understand the scope of variables. If you are reading a math book then 'u' on the second page might mean the same thing as 'u' on the fifteenth page or it might mean something entirely different. The meaning of 'u' might even change from paragraph to paragraph or from sentence to sentence.

    As an advanced topic, another trick problem is

    Given
    [itex] f(A,B) = 6A + B [/itex]

    Find [itex] \frac{\partial f}{\partial x} [/itex]

    A traditional interpretation of [itex] \frac{\partial f}{\partial x} [/itex] is that it means "the partial derivative of [itex] f [/itex] with respect to its first argument". With that interpetation the answer would be [itex] \frac{\partial f}{\partial x} = 6. [/itex] But it requires that kind of "cultural" understanding to make any sense of the problem.
     
    Last edited: Jul 1, 2013
  8. Jul 1, 2013 #7

    Mark44

    Staff: Mentor

    My understanding is that the prime notation is a slight change from the dot notation that Newton used. Newton was concerned exclusively with functions in which time was the independent variable, so a derivative was always with respect to time.
     
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