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Chain rule explanation

  1. Jan 7, 2012 #1
    Hello,

    Looking through a book on calculus I found the following explanation for the chain rule and I have one unclear thing that I'd like to ask for help on.

    The canonical example is used, y is a function of u: [tex]y = u^{n}[/tex] and u is a function of x (let's say) [tex]u = 3x - 2[/tex] therefore by composition y is a function of x.

    By the definition of the derivative:

    [tex]\frac{dy}{dx} = \lim_{\Delta x\to0}{\frac{f(x+\Delta x) - f(x)}{\Delta x}}[/tex]
    we can say that
    [tex]\frac{dy}{dx} = \lim_{\Delta x\to0}{\frac{\Delta y}{\Delta x}}[/tex]

    in the same way that
    [tex]\frac{dy}{du} = \lim_{\Delta u\to0}{\frac{\Delta y}{\Delta u}}[/tex]
    and
    [tex]\frac{du}{dx} = \lim_{\Delta x\to0}{\frac{\Delta u}{\Delta x}}[/tex]

    Then, using the equation:
    [tex]\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}[/tex]

    we can write

    [tex]\frac{dy}{dx} = \lim_{\Delta x\to0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x\to0}(\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}) = \lim_{\Delta x\to0}\frac{\Delta y}{\Delta u}\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}[/tex]

    then it is said that "However because u is a function of x, [tex]\Delta x \rightarrow 0[/tex] exactly when [tex]\Delta u \rightarrow 0[/tex] so

    [tex]\lim_{\Delta x\to0}\frac{\Delta y}{\Delta u} \lim_{\Delta x\to0}\frac{\Delta u}{\Delta x} = \lim_{\Delta u\to0}\frac{\Delta y}{\Delta u}\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x} = \frac{dy}{du} \cdot \frac{du}{dx}[/tex]

    And so the chain rule.

    The part in red is the one I couldn't understand, specifically why does [tex]\Delta x \rightarrow 0[/tex] exactly when [tex]\Delta u \rightarrow 0[/tex] ?

    Sorry for the lengthy post and thanks in advance for any help.

    Cheers!
     
  2. jcsd
  3. Jan 7, 2012 #2

    micromass

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    Buy another, more rigorous book.

    "Proofs" like this want to make me rip my hair out.
     
  4. Jan 7, 2012 #3
  5. Jan 7, 2012 #4

    Yes, "u" is a function of "x". This implies that u is dependent on x and x is
    an independent variable. Now since u depends on x, any change in x
    produces a change in u. But if x does not change, there will be no
    "CHANGE" in the dependent variable u.
    :)
     
  6. Jan 7, 2012 #5
    Or,
    Δu = Δf(x)
    Δu = [3*(x+Δx) - 2] - [3*x - 2]
    Δu = 3*Δx
    now if Δx→0
    then Δu→0.

    Hope this helps.
    :)
     
  7. Jan 7, 2012 #6

    Fredrik

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    That particular ##\Delta u## is defined by ##\Delta u=u(x+\Delta x)-u(x)##. Clearly ##\Delta x=0## implies ##\Delta u=0##. The converse implication is explained here. However, I agree with micromass. This isn't the way to prove the chain rule. The problem is that ##\Delta u## is defined as I said when it appears in a numerator but not when it appears in a denominator.

    The Wikipedia page has some short proofs, but I think I still prefer to prove it directly from the definitions, without the tricks that are used in the Wikipedia proofs. Unfortunately such a proof is about 2-3 pages long if all the details are explained.
     
    Last edited: Jan 7, 2012
  8. Jan 7, 2012 #7
    @micromass
    @spamiam
    Thanks for the replies, I've seen other proofs which I've been able to workout on paper and those make sense, I just got really hung up on this specific 'proof' because I couldn't make sense of the red part.

    @sahil_time
    @Fredrik
    Thank you! I though that the 'proof' presented in the book was for the general case and it didn't make sense so I didn't bother to work it out on paper, I can see now it was strictly (well .. more or less) related to the example provided; sorry for wasting your time, it's clear now!

    So with that being said, my problem is resolved, the input was great, thanks a lot!

    Cheers!
     
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