Find the derivative of $f$ from the first principle,where $f$ is given by $f(x) = x + \frac{1}{x}$.

The function is not defined at $x = 0$. Using the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{(x+h + \frac{1}{x+h}) - (x + \frac{1}{x})}{h}$
$= \lim_{h \to 0} \frac{1}{h} [h + \frac{1}{x+h} - \frac{1}{x}]$
$= \lim_{h \to 0} \frac{1}{h} [h + \frac{x - (x+h)}{x(x+h)}]$
$= \lim_{h \to 0} \frac{1}{h} [h - \frac{h}{x(x+h)}]$
$= \lim_{h \to 0} [1 - \frac{1}{x(x+h)}]$
$= 1 - \frac{1}{x^2}$
Thus,the derivative is $1 - \frac{1}{x^2}$ for $x \neq 0$.