Differentiation: How does it overcome the 0/0 problem?

[cscott]

How does differentiation overcome the fact that as we let $\Delta x \rightarrow 0$ until it is truly 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]

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]

How does differentiation overcome the fact that as we let $\Delta x \rightarrow 0$ until it is truly 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 [itex]x+\Delta{x}[/tex]<br /> <br /> I'm kind of confused on your question. If you could explain your thoughts more, perhaps we could help you more.[/itex]

 

[Hurkyl]

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.

 

[quantumdude]

How does differentiation overcome the fact that as we let $\Delta x \rightarrow 0$ until it is truly 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...

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?

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

 

[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]

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.