# 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:

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

To this:

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

2. First we need a simple theorem.

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

Now we define a derivative.

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

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

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

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

Proof: The proof is nearly identical to Theorem 1.

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

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