Differentiation: Magic?

cscott

How does differentiation overcome the fact that as we let $\Delta x \rightarrow 0$ until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of $\frac{0}{0}$?

quetzalcoatl9

$$\lim_{\Delta x \rightarrow 0} \Delta x = 0 \neq dx$$

we are considering the function $$\frac{dy}{dx}$$

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

is different than the limit of it's parts.

robphy

Homework Helper
Gold Member
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 $$\Delta f$$ (which depends on $$\Delta x$$) and $$\Delta x$$ itself as $$\Delta x$$ gets small.

Jameson

cscott said:
How does differentiation overcome the fact that as we let $\Delta x \rightarrow 0$ until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of $\frac{0}{0}$?
They aren't the exact same coordinate. Hence the $x+\Delta{x}[/tex] I'm kind of confused on your question. If you could explain your thoughts more, perhaps we could help you more. Hurkyl Staff Emeritus Science Advisor Gold Member The question you need to be asking yourself is "what is a limit?" Once you have that down, it will be (more) clear how the definition of a derivative in terms of a limit works. Tom Mattson Staff Emeritus Science Advisor Gold Member cscott said: How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0$ until it is truely 0, our two coordinates to take the slope with will be the same and thus give a slope of $\frac{0}{0}$?
To expand on Jameson's comment....

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

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

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

So in the definition of the derivative:

$$f'(x)=\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$,

we can see that it is $|\Delta x - 0|$ that is forever restricted to be above zero.

edit:

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

$\lim_{x \rightarrow c}f(x)=L$

Last edited:

Jameson

I think "itex" is limited to one line... so the the x ->c can't fit under.

cscott

Thanks for the new insight. I'm going to read more on limits and such.

mathwonk

Homework Helper
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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