Differentiation: Magic?

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How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of [itex]\frac{0}{0}[/itex]?
 
[tex]\lim_{\Delta x \rightarrow 0} \Delta x = 0 \neq dx[/tex]

we are considering the function [tex]\frac{dy}{dx}[/tex]

and the limit of the whole thing, [tex]\lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}[/tex]

is different than the limit of it's parts.
 

robphy

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Try f(x)=x^2 in quetzalcoatl9's formula (the definition of the derivative).
Expand out the numerator. Do some algebra.
Lastly, take the limit.
See what you get.

Try x^3, etc... sin(x),exp(x), etc...

In some sense what you're doing is measuring the relative rate between [tex] \Delta f[/tex] (which depends on [tex]\Delta x[/tex]) and [tex]\Delta x[/tex] itself as [tex]\Delta x[/tex] gets small.
 
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cscott said:
How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of [itex]\frac{0}{0}[/itex]?
They aren't the exact same coordinate. Hence the [itex]x+\Delta{x}[/tex]

I'm kind of confused on your question. If you could explain your thoughts more, perhaps we could help you more.
 

Hurkyl

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The question you need to be asking yourself is "what is a limit?" Once you have that down, it will be (more) clear how the definition of a derivative in terms of a limit works.
 

Tom Mattson

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cscott said:
How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of [itex]\frac{0}{0}[/itex]?
To expand on Jameson's comment....

Jameson said:
They aren't the exact same coordinate. Hence the [itex]x + \Delta x[/itex]
cscott, go back to the definition of a limit, which says:

Let [itex](a,b)[/itex] be an open interval containing [itex]c[/itex]. Let f be a function defined on [itex](a,b)[/itex], except possibly at [itex]c[/itex].

[itex]\lim_{x \rightarrow c}f(x)=L[/itex] iff for every [itex]\epsilon > 0[/itex], there exists [itex]\delta > 0[/itex] such that [itex]0<|x-c|<\delta[/itex] implies that [itex]|f(x)-L|<\epsilon[/itex].
Note that the second inequality really implies that [itex]0 \leq |f(x)-L| < \epsilon[/itex], which means that [itex]f(x)-L[/itex] can indeed equal zero. But in the first inequality, [itex]|x-c|[/itex] is forever restricted to be above zero.

So in the definition of the derivative:

[tex]f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex],

we can see that it is [itex]|\Delta x - 0|[/itex] that is forever restricted to be above zero.

edit:

Hmmm....Why didn't LaTeX format this right?

[itex]\lim_{x \rightarrow c}f(x)=L[/itex]
 
Last edited:
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I think "itex" is limited to one line... so the the x ->c can't fit under.
 
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Thanks for the new insight. I'm going to read more on limits and such.
 

mathwonk

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or you could take the classical way out, of descartes etc.

i.e. when f is a polynomial, then f(x)-f(a) = [x-a][g(x)] for some factor g.

then you are asking essentially for the value of the quotient [f(x)-f(a)]/(x-a)

= [(x-a)g(x)]/(x-a) at x=a.

obviously you should factor out the x-a first, then the answeer is g(a), which has nothing to do with the fact that you get 0/0 before you factor out.


now in cases where you do not know how to factor, you plug in a sequence of numbers for x that are merely close to a, i.e. you let x = a+e for small e.

then you get the sequence of approximations
[(e) g(a + e)]/e = g(a+e). as e gets smaller and smaller, hopefully you are eventually able to guiess what g(a) should be from all these approximations of form g(a+e).

if so you have "taken the limit" as x goes to a.
 

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