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Understanding differentials and differentiation

  1. Aug 18, 2009 #1
    I am very interested in math and I find calculus to be a particularly interesting subject, but one major problem I have with it is that I cannot find a consistent explanation of the rules of differentials (infantesimals) that explains all the things mathematicians do with them. I have truly looked everywhere, even in my expensive calculus books, but they just start using the notation and expect it to somehow make sense.

    Therefore, I will try to explain my understanding of them and I would like you to point out where my errors are, since some of my understanding contradicts credible sites like wolfram:

    First to develop some tools:

    The differential is written

    [tex]dx[/tex]

    where x represents the variable that is to be varied by an infantesimaly small amount.

    Another way to write a differential, usually when the variable or quantity to be varied is an expression is

    [tex]d(x)[/tex]

    where [tex]x[/tex] is the expression in question.

    The operation of differentiation is NOT the same thing as finding the derivative (as it is defined on Wolfram and on Wikipedia) as a derivative is not a differential but a RATIO of differentials.

    The derivative can be more accurately defined as first differentiation then dividing by the differential of the variable you want to find change with respect to.

    For example, applying the operator [tex]\frac{d}{dx}[/tex] does not mean to differentiate, but rather to differentiate THEN divide by [tex]dx[/tex].

    For example when given [tex]y = x[/tex] and asked to differentiate the correct answer is

    [tex]dy = dx[/tex]

    and when asked to find the derivative with respect to x this means first differentiate:

    [tex]dy = dx[/tex]

    then divide both sides by the differential of [tex]x[/tex]

    [tex] \frac{dy}{dx} = \frac{dx}{dx}[/tex]

    [tex]
    \frac{dy}{dx} = 1
    [/tex]
    where the quantity [tex]\frac{dy}{dx}[/tex] is referred to as the 'derivative'.

    The power rule is a special case of the chain rule and a distinction is made between them only because math professors are evil and wish to confuse students by making things seem more complicated:

    [tex]y = (x)^n[/tex]

    [tex]dy = n(x)^{n-1}dx[/tex]

    If [tex]x[/tex] was a more complicated expression, as it is in the chain rule, we simply use direct substitution:

    Say if [tex]x = 3u[/tex].

    [tex]dy = n(3u)^{n-1}d(3u)[/tex]

    [tex]dy = n(3u)^{n-1}(3)du[/tex]

    Product rule:

    [tex]y = f(x)g(x)[/tex]

    [tex]dy = f'(x)(dx)g(x) + f(x)g'(x)(dx)[/tex]

    Quotient rule:

    [tex]y = \frac{f(x)}{g(x)}[/tex]

    [tex]dy = \frac{g(x)f'(x)(dx) - f(x)g'(x)(dx)}{(g(x))^2}[/tex]

    Differentiating a constant is zero:

    [tex]d(c) = 0[/tex]

    Differentiating a differential is described in notation as raising the d to higher and higher powers:

    For example repeatedly differentiating y we get:

    [tex]y[/tex]

    [tex]dy[/tex]

    [tex]d^{2}y[/tex]

    [tex]d^{3}y[/tex]



    The notation for a second derivative is

    [tex]\frac{d^2y}{{dx}^2}[/tex]

    This can be understood better as

    [tex]\frac{d}{dx}\frac{dy}{dx}[/tex]

    Here is a proof of this fact:

    First differentiate dy/dx then divide by the quantity dx:

    [tex]\frac{d}{dx}\frac{dy}{dx} = d{\frac{dy}{dx}}{\frac{1}{dx}}[/tex]

    using the quotient rule:

    [tex]d{\frac{dy}{dx}}[/tex] = low d high less high d low, underneath denominator squared will go =

    [tex]\frac{(dx)(d^{2}y) - (dy)(d^{2}x)}{{dx}^2}[/tex]

    Break this up into

    [tex]\frac{d^2y}{dx} - \frac{(dy)(d^2x)}{{dx}^2}[/tex]

    we can note that

    [tex]\frac{(dy)(d^2x)}{(dx)^2} = 0[/tex]

    because

    [tex]\frac{(dy)(d^2x)}{(dx)^2} = (dy)(\frac{d}{dx})(\frac{dx}{dx})[/tex]

    and since

    [tex]\frac{dx}{dx} = 1[/tex]

    [tex](dy)(\frac{d}{dx})(1) = 0[/tex]

    since [tex]d(1) = 0[/tex]

    Now that we know

    [tex]d(\frac{dy}{dx}) = \frac{(d^2y)}{dx} - \frac{(dy)(d^2x)}{(dx)^2} = \frac{(d^2y)}{dx}[/tex]

    we can say that

    [tex](\frac{d}{dx})(\frac{dy}{dx}) = d(\frac{dy}{dx})(\frac{1}{dx}) = (\frac{d^2y}{dx})(\frac{1}{dx}) = \frac{d^2y}{{dx}^2}[/tex]

    which is the standard notation for the second derivative.
     
    Last edited: Aug 19, 2009
  2. jcsd
  3. Aug 18, 2009 #2
    Just consult any calculus book for the Leibniz notation and usually the inside front cover will contain a much better printing of that mess you posted above.

    Thanks
    Matt
     
  4. Aug 18, 2009 #3
    Unfortunately my calculus book only has a short paragraph describing Leibniz notation and only uses them to define the derivative and does not give me a confident understanding of how to operate with differentials.

    I may return later and use the proper BBcodes to describe the mess I have posted above more clearly.
     
  5. Aug 18, 2009 #4
    If you want to understand why and what is happening, you should start from the definition of limits, not with the differentials. Apply the definition of limit to each derivation rule and you'll see it.

    A good reference to grasp these concepts the book on Calculus by Gilbert Strang is a good choice, which is freely available from MIT.

    http://ocw.mit.edu/ans7870/resources/Strang/strangtext.htm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Aug 19, 2009 #5
    I believe part of the confusion you may be encountering is that "differentiation" is used with slightly different meanings.

    Here are some "differentiations"

    [itex]d=(x \mapsto dx)[/itex] is the general differential operator.

    [itex]\frac{d}{dx}=(y \mapsto y')[/itex] is differentiation with respect to x and yields the derivative (in x).

    [itex]\frac{d}{dt}= \left( f(u) \mapsto f'(u) \cdot \frac{du}{dt} \right)[/itex] is differentiation with respect to t and frequently appears when discussing "implicit differentiation."

    Finding a derivative is a specific kind of differentiation (with respect to a variable).

    I realize this may not really make the subject easier to digest, but I hope I made something slightly more clear.

    --Elucidus
     
  7. Aug 19, 2009 #6
    Thank you for your responses. I have changed the formatting to make what I wrote more readable.

    I would like to know what parts of what I have written above would be considered invalid on an exam so I can learn them correctly or not use them. Mainly, I want to know if my proof of the second derivative notation is valid because it seems to work so perfectly.

    I understand how limits relate to differentials but I guess I'm looking for an axiomatic set of rules that define what I can and can't do with differentials without thinking about what they are intrinsically (maybe no such list of rules exists). I believe the study of non-standard analysis may explain these concepts, does it not?
     
  8. Aug 22, 2009 #7
    Offhand and to my unskilled/trained eyes, nothing looks inaccurate.
     
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