# Thread: Second Derivite as a difference quitient?

1. ## Second Derivite as a difference quitient?

Okay, I was wondering if there was a way to take a sortcut to the second derivitive of a function? What I'm talking about here directly involves the differnce quotient. I know all the little rules of differentiation that are derived from the difference quotient, and I know how to use them to find the second derivitve pretty easily. But I'm talking about something more "theoretical", and less "applicable". Let me start by using this definition:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Now, lets say we are not going to use any of the fancy simple shortcuts that have been proved about diferentiation; lets say that we are required to use the difference quotient, can we find a short cut some how? Heres what I mean:

$\displaystyle f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h}$

Now, expanding things gives:

$\displaystyle f''(x) = \lim_{h \to 0} \frac{\lim_{h \to 0} \frac{f((x+h)+h) - f(x+h)}{h} - \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}}{h}$

My question is, can we algebraically manipulate the expression above to get a different looking difference quotient? Or some other sort of limit expression that would be a 'short-cut' to the second derivitive? Some function involving $\displaystyle f(x)$ that we take the limit of as $\displaystyle h \to 0$? (I'm gona call this function $\displaystyle g(x)$). So that we can end up with a simpler form of the limit above that takes us striaght to the second derivitive? Something of the form:

$\displaystyle f''(x) = \lim_{h \to 0} g(x)$

?

2. Hi,

I know something that is not far from what you ask :

under certain circumstances, $\displaystyle f''(x) = \lim_{h \to 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2}$, but it is not the same exactly the same thing.
In french, we call it "pseudo dérivée seconde", but I don't know how it is called in english...

3. Originally Posted by mfetch22
...$\displaystyle f''(x) = \lim_{h \to 0} \frac{\lim_{h \to 0} \frac{f((x+h)+h) - f(x+h)}{h} - \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}}{h}$...
I don't think there's a simple expression that always works, but you've hit upon something important in mathematics. Pretty much all of the formulas you see with dummy variables assume the variables don't already appear on the inside. Since you have h's for the regular derivative in there, the outer limit can't use h. You have to write something like $\displaystyle f''(x) = \lim_{k \to 0} \frac{\lim_{h \to 0} \frac{f((x+k)+h) - f(x+k)}{h} - \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}}{k}$.