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

In summary, differentiation is able to overcome the fact that the two coordinates used to calculate the slope will eventually be the same and result in a slope of 0/0 by considering the limit of the function as \Delta x approaches 0. This is done by measuring the relative rate between \Delta f and \Delta x as \Delta x gets smaller. The definition of a limit is key in understanding how the derivative in terms of a limit works. Alternatively, you can also use the classical method of factoring to find the derivative by plugging in a sequence of numbers close
  • #1
cscott
782
1
How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truly 0, our two coordinates to take the slope with will be the same and thus give a slope of [itex]\frac{0}{0}[/itex]?
 
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  • #2
[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.
 
  • #3
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.
 
  • #4
cscott said:
How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truly 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.
 
  • #5
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.
 
  • #6
cscott said:
How does differentiation overcome the fact that as we let [itex]\Delta x \rightarrow 0[/itex] until it is truly 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:
  • #7
I think "itex" is limited to one line... so the the x ->c can't fit under.
 
  • #8
Thanks for the new insight. I'm going to read more on limits and such.
 
  • #9
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.
 

1. What is differentiation?

Differentiation is the process of finding the rate of change of a function. It involves calculating the slope of a curve at a specific point.

2. Why is differentiation important in science?

Differentiation is important in science because it allows us to understand how a system changes over time. It is used extensively in fields such as physics, chemistry, biology, and economics to model and analyze real-world phenomena.

3. What is the difference between differentiation and integration?

Differentiation and integration are inverse operations. Differentiation finds the slope of a curve at a specific point, while integration finds the area under the curve. In other words, differentiation is a process of finding derivatives, while integration is a process of finding antiderivatives.

4. What is the chain rule in differentiation?

The chain rule is a fundamental rule in differentiation that allows us to find the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

5. How is differentiation used in real-life applications?

Differentiation is used in real-life applications to solve various problems related to rates of change. For example, it is used in economics to optimize production and profit, in physics to analyze motion and forces, and in engineering to design and improve systems. It is also used in data analysis to find relationships between variables and make predictions.

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