Differentiability on an interval.

Definition 6.4.1   The function $f$ is differentiable on the open interval $I$ if it is differentiable at each point of $I$ .

Definition 6.4.2   The function $f$ is differentiable on the closed interval $[a,b]$ if it verifies the following conditions:

1. $f$ is differentiable on the open interval $(a,b)$ ;
2. $f$ is differentiable on the right at $a$ ;
3. $f$ is differentiable on the left at $b$ .

There are similar definitions for other intervals, like $[a,+\infty)$ , etc.

Example 6.4.3   Take $f(x)=\sqrt{x}$ .

For $x_0 \neq 0$ , $\frac {\sqrt{x}-\sqrt{x_0}}{x-x_0}=\frac {1}{\sqrt{x}+\sqrt{x_0}} \Longrightar... ...et{x \rightarrow 0}{\text{lim}} \frac {f(x)-f(0)}{x-0}= \frac {1}{2 \sqrt{x_0}}$ .

Hence, the square root function is differentiable on $(0,+\infty)$ .

Take now $x_0=0$ . Then: $\frac {\sqrt{x}-\sqrt{0}}{x-0} = \frac {1}{\sqrt{x}} \Longrightarrow \underset... ...derset{>}{\rightarrow} 0}{\text{lim}} \frac {\sqrt{x}-\sqrt{0}}{x-0} = +\infty.$

The square root function is not differentiable on the right at 0.

Theorem 6.4.4 (Table of usual derivatives)       

$f(x)$	$f'(x)$	conditions
$k$	0
$x$	1
$x^{\alpha}$	$\alpha x^{\alpha -1}$	$\alpha \in \mathbb{Q}$ , $x>0$ when $\alpha \not\in \mathbb{Z}$
$\sqrt{x}$	$\frac {1}{2\sqrt{x}}$	$x>0$
$\ln x$	$\frac 1x$	$x>0$
$e^x$	$e^x$
$\cos x$	$-\sin x$
$\sin x$	$\cos$ x
$\tan x$	$\frac {1}{\cos^2 x}=1+\tan^2 x$	$x \in \mathbb{R}- \left\{ \frac {\pi}{2} + k \mathbb{Z} \right\}$
$\cosh x$	$\sinh x$
$\sinh x$	$\cosh x$
$\tanh x$	$\frac {1}{\cosh^2 x}=1-\tanh^2 x$