One-sided derivative at one point.

Definition 6.6.1   The function $ f$ id differentiable on the left (resp. on the right) at $ x_0$ if the quotient $ \frac {f(x)-f(x_0)}{x-x_0}$ has a finite limit on the left (resp. on the right) at $ x_0$ . This limit is called the first derivative on the left (resp. on the right) of $ f$ at $ x_0$ and is denoted by $ f'_l(x_0)$ (resp. $ f'_r(x_0)$ ).

\fbox{
$f'_l(x_0)=\underset{x \underset{<}{\rightarrow} x_0}{\text{lim}} \frac {...
...)}{x-x_0}
=\underset{h \rightarrow 0}{\text{lim}} \frac {f(x_0+h)-f(x_0)}{h}$
} \fbox{
$f'_r(x_0)=\underset{x \underset{>}{\rightarrow} x_0}{\text{lim}} \frac {...
...)}{x-x_0}
=\underset{h \rightarrow 0}{\text{lim}} \frac {f(x_0+h)-f(x_0)}{h}$
}

Example 6.6.2   Take $ f(x)=\vert x\vert$ . Then:

Proposition 6.6.3   The function $ f$ is differentiable at $ x_0$ if, and only if, it verifies the three following conditions:
  1. $ f$ is differentiable on the left at 0;
  2. $ f$ is differentiable on the right at 0;
  3. $ f'_l(x_0)=f'_r(x_0)$ .

Example 6.6.4   The absolute value function is not differentiable at 0. We showed in the previous example that this function is differentiable on the left and differentiable on the right at 0, but as the one-sided derivatives are different, the function is not differentiable at 0.

Example 6.6.5   Take $ f(x)=x\vert x\vert$ . Then: Therefore, $ f$ is differentiable at 0 and $ f'(0)=0$ .

Noah Dana-Picard 2007-12-28