# Thread: Definition of the Derivative

1. ## Definition of the Derivative

I am going over some of my notes, and I was taking a different way to get to the definition.

Find the slopes between two points, x - h AND x + h. We want the "slope" or derivative at x, so set the limit as h approaches zero. How do I use algebra to go from this:

$lim_{h\rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

To this:

$lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

2. First we need a simple theorem.

Theorem: Suppose $g$ is a function defined on an open interval containing $a$ (except possibly at $a$) so that $\lim_{x\to a} g(x) = L$ where $L$ is a real number. Suppose $f$ is a function defined on an open interval containing $L$ so that $f$ is continous at $L$ then $f\circ g$ is defined in an open interval containing $L$ (except possibly at $L$) and $\lim_{x\to a}f(g(x)) = f(L)$.

Now we define a derivative.

Definition: Suppose $f$ is a function defined on an open interval $I$ containing $a$. Define the function $g:I\setminus \{ a\} \mapsto \mathbb{R}$ as $g(x) = \frac{f(x)-f(a)}{x-a}$ then $g$ is defined on an open interval containing $a$ (except at $a$). If it happens that $\lim_{x\to a} g(x)$ exists (real number) then we say $f$ is differenciable at $a$ and we call this limit $f'(a)$.

Theorem 1: Suppose that $f$ is differenciable at $a$ defined on an open interval $I$ containing $a$. Define the function $g(x) = \frac{f(x+a) - f(a)}{x}$ for all $x$ such that $x+a\in I \mbox{ and }x\not = 0$ then clearly $g$ is defined on an open interval containing $0$ (except at at $0$) and $\lim_{x\to 0}g(x) = f'(a)$.

Proof: The part where $g$ is defined around $0$ is straightforward. Now to show $\lim_{x\to 0}g(x) = f'(a)$ we use the theorem above. Let $h(x) = x + a$ and let $F(x) = \frac{f(x)-f(a)}{x-a}$ then the conditions of the theorem are satisfied and $f'(a)= \lim_{x\to 0}F(h(x)) = \lim_{x\to a}\frac{f(x+a) - f(a)}{x}$.

Theorem 2: Suppose that $f$ is differenciable at $a$ defined on an open interval $I$ containing $a$. Define the function $g(x) = \frac{f(a) - f(x-a)}{x}$ for all $x$ such that $x-a\in I \mbox{ and }x\not = 0$ then clearly $g$ is defined on an open interval containing $0$ (except at $0$) and $\lim_{x\to 0}g(x) = f'(a)$.

Proof: The proof is nearly identical to Theorem 1.

Corollarly: Suppose that $f$ is differenciable at $a$ defined on an open interval $I$ containing $a$. Define the function $g(x) = \frac{f(x+a) - f(x-a)}{2x}$ for all $x$ such that $x\pm a \in I \mbox{ and }x\not = 0$ then clearly $g$ is defined on an open interval containing $0$ (except at $0$) and $\lim_{x\to 0}g(x) = f'(a)$.

Proof: Note that $\frac{f(x+a) - f(x-a)}{2x} = \frac{1}{2}\cdot \frac{f(x+a) - f(a)}{x} + \frac{1}{2} \cdot \frac{f(a) - f(x-a)}{x}$. This means $\lim_{x\to 0}\frac{f(x+a)-f(x-a)}{2x} = \frac{1}{2}f'(a) + \frac{1}{2}f'(a) = f'(a)$ by the above theorems.