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6 Comments

  1. Anitej Banerjee
    June 26, 2016 @ 3:11 am

    Wow, that really was mind blowing!
    I did come across a few hitches in the logic — perhaps due to my own misunderstandings of the topic.
    Firstly, why define g'(c) to be the lim (x->c) of [g(x) – g(c)]/[x-c].
    If you were to follow the definition from most textbooks:
    f'(x) = lim (h->0) of [f(x+h) – f(x)]/[h]
    Then, for g'(c), you would come up with:
    g'(c) = lim (h->0) of [g(c+h) – g(c)]/[h]
    Perhaps the two are the same, and maybe it’s just my loosey-goosey way of thinking about the limits that is causing this confusion…
    Secondly, I don’t understand how bold Q(x) works. I understand the law of composite functions limits part, but it just seems too easy — just defining Q(x) to be f'(x) when g(x) = g(c)… I can’t pin-point why, but it feels a little bit like cheating :P.

    Lastly, I just came up with a geometric interpretation of the chain rule — maybe not so fancy :P.
    f(g(x)) is simply f(x) with a shifted x-axis [Seems like a big assumption right now, but the derivative of g takes care of instantaneous non-linearity]. g'(x) is simply the transformation scalar — which takes in an x value on the g(x) axis and returns the transformation scalar which, when multiplied with f'(x) gives you the actual value of the derivative of f(g(x)). I like to think of g(x) as an elongated x axis/input domain to visualize it, but since the derivative of g'(x) is instantaneous, it takes care of the fact that g(x) may not be as linear as that — so g(x) could also be an odd-powered polynomial (covering every real value — loved that article, by the way!) but the analogy would still hold (I think).

    Once again, thank you very much! 😀

    Reply

  2. Math Vault
    June 26, 2016 @ 2:46 pm

    Hi Anitej. For the first question, the derivative of a function at a point can be defined using both the x-c notation and the h notation. In fact, using a stronger form of limit comparison law, it can be shown that if the derivative exists, then the derivative as defined by both definitions are equivalent.

    For the second question, the bold Q(x) basically attempts to patch up Q(x) so that it is actually continuous at g(c). Now, if we define the bold Q(x) to be f'(x) when g(x)=g(c), then not only will it not take care of the case where the input x is actually equal to g(c), but the desired continuity won’t be achieved either.

    And as for the geometric interpretation of the Chain Rule, that’s definitely a neat way to think of it!

    Reply

    • Anitej Banerjee
      June 27, 2016 @ 5:06 am

      Well that sorts it out then… err, mostly.
      But why resort to f'(c) instead of f'(g(c)), wouldn’t that lead to a very different value of f'(x) at x=c, compared to the rest of the values [That does sort of make sense as the limit as x->c of that derivative doesn’t exist]?
      Either way, thank you very much — I certainly didn’t expect such a quick reply! 🙂

      Reply

  3. Math Vault
    June 27, 2016 @ 3:40 pm

    Oh. It is f'[g(c)]. Remember, g being the inner function is evaluated at c, whereas f being the outer function is evaluated at g(c). In particular, the focus is not on the derivative of f at c. You might want to go through the Second Attempt Section by now and see if it helps.

    Reply

  4. Pranjal
    November 8, 2016 @ 1:20 am

    Thank you. This is awesome . This is one of the most used topic of calculus . You have explained every thing very clearly but I also expected more practice problems on derivative chain rule.

    Reply

    • Math Vault
      November 8, 2016 @ 7:20 pm

      Hi Pranjal. For calculus practice problems, you might find the book “Calculus” by James Stewart helpful. It’s under the tag “Applied College Mathematics” in our resource page.

      Reply

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