Differentiability at one point.

Definition 6.1.1   Let $ f$ be a function defined on a neighborhood of $ x_0$ . The function $ f$ is differentiable at $ x_0$ if the quotient $ \frac {f(x)-f(x_0)}{x-x_0}$ has a finite limit for $ x$ arbitrary close to $ x_0$ . This limit is called the first derivative of $ f$ at $ x_0$ and is denoted by $ f'(x_0)$ .

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

Example 6.1.2       

Let $ f(x)=x^2$ and $ x_0=1$ . Then we have:

$\displaystyle \frac {f(x)-f(x_0)}{x-x_0} = \frac {x^2-1}{x-1} = x+1$    

Thus:

$\displaystyle \underset{x \rightarrow 1}{\text{lim}} \frac {f(x)-f(1)}{x-1}=\underset{x \rightarrow 1}{\text{lim}} (x+1)=2$    

We proved that $ f$ is differentiable at 1 and that $ f'(1)=2$ .

Example 6.1.3       

Let $ f(x)=1/x$ and $ x_0=2$ . Then we have:

$\displaystyle \frac {f(x)-f(2)}{x-2} = \frac {\frac 1x - \frac {1}{2}}{x-2} =\frac {\frac {2-x}{2x}}{x-2} =-\frac {-1}{2x}$    

Thus

$\displaystyle \underset{x \rightarrow 2}{\text{lim}} \frac {f(x)-f(2)}{x-2} = \underset{x \rightarrow 2}{\text{lim}} \left( -\frac {-1}{2x} \right)= -\frac 14.$    

We proved that $ f$ is differentiable at 2 and that $ f'(2)=-1/4$ .

Definition 6.1.4   Sometimes, we denote:

$\displaystyle f'(x_0)=\frac {df}{dx} \bracevert_{x=x_0} =\frac {df}{dx} \bracevert_{x_0}$    

Noah Dana-Picard 2007-12-28