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$     

Noah Dana-Picard 2007-12-28