One sided derivatives (left hand and right hand derivatives)

Definition 10.2's limit is two-sided — $\Delta x\to 0$ from both directions at once must give the same value. Splitting it into its two one-sided pieces is what lets us pin down exactly where and why a derivative can fail to exist, and is essential at the endpoints of a closed interval, where the function isn't even defined on one side.

For $y=f(x)$ defined on an open interval $(a,b)$ containing $x_0$, the left hand derivative $f'(x_0^-)$ and right hand derivative $f'(x_0^+)$ are defined by

provided the respective limits exist. Exactly as for two-sided limits generally, the full derivative

exists if and only if both $f'(x_0^-)$ and $f'(x_0^+)$ exist and are equal: $f'(x_0^-)=f'(x_0^+)$. If even one of the two one-sided derivatives fails to exist, or they exist but disagree, then $f$ is not differentiable at $x_0$ — this single fact is the working test used throughout §10.3 and Exercise 10.1.

Writing $h=\Delta x$ (with $h>0$ understood to shrink to $0$ from the appropriate side) gives the equivalent, often more convenient forms

Differentiability on a closed interval.

Definition 10.3. $f$ is differentiable on $[a,b]$ if it is differentiable on the open interval $(a,b)$, and, at the two endpoints,

$f'(a) = \lim_{\Delta x\to 0^+}\frac{f(a+\Delta x)-f(a)}{\Delta x} = \lim_{h\to0}\frac{f(a+h)-f(a)}{h},\ h>0,$

$f'(b) = \lim_{\Delta x\to 0^-}\frac{f(b+\Delta x)-f(b)}{\Delta x} = \lim_{h\to0}\frac{f(b-h)-f(b)}{h},\ h>0.$

That is, at the left endpoint $a$ only the right-hand limit makes sense (there is no function to the left of $a$), and at the right endpoint $b$ only the left-hand limit makes sense. …