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 $f'(x_0)$ .

Example 6.1.2       

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

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

Thus:

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

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:

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

Thus

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

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

Definition 6.1.4   Sometimes, we denote:

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