I need to give a definition of the derivative from right and from left. (I know it doesn't make a whole lot of sense, but it is supposed to be similar to the same thing as the limits from the right and what not because the definition of the derivative is just the limit of $\frac{f(x)-f(a)}{x-a}$)
I have this:
the derivative from the right of $f$ at $a$ is $\lim_{x\to a+}=\frac{f(x)-f(a)}{x-a}$ if for every $\epsilon >0$, there exists a $\delta> 0$, such that if $0<x-a<\delta$, then $|f(x)-f'(a)|<\epsilon.$
Is this at all right?
$$\text{Right Derivative: }\partial_+f(a):=\lim_{{\scriptstyle x\to a+\atop\scriptstyle x\in I}}\frac{f(x)-f(a)}{x-a}\\ \implies \forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - a| < \delta \ \Rightarrow \ \left|\frac{f(x)-f(a)}{x-a} - \partial_+f(a)\right| < \varepsilon)\text{ as $x\to a$ from the right}.$$ $$\text{Left Derivative: }\partial_-f(a):=\lim_{{\scriptstyle x\to a-\atop\scriptstyle x\in I}}\frac{f(x)-f(a)}{x-a}\\ \implies \forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - a| < \delta \ \Rightarrow \ \left|\frac{f(x)-f(a)}{x-a} - \partial_-f(a)\right| < \varepsilon)\text{ as $x\to a$ from the left}.$$ (Also, you may define derivatives acting to the left or right as $$f \stackrel{\leftarrow }{\partial }_x g = \frac{\partial f}{\partial x} \cdot g\\ f \stackrel{\rightarrow }{\partial }_x g = f \cdot \frac{\partial g}{\partial x}.)$$