Find from first principles, the derivative with respect to \( x \) of the function \( f: x \rightarrow 3 x+\frac{1}{x} \).

To find the derivative of the function \( f(x) = 3x + \frac{1}{x} \) from first principles, we start with the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Now, let's compute \( f(x+h) \): \[ f(x+h) = 3(x+h) + \frac{1}{x+h} = 3x + 3h + \frac{1}{x+h} \] Now, substituting \( f(x+h) \) and \( f(x) \) into the derivative formula: \[ f'(x) = \lim_{h \to 0} \frac{(3x + 3h + \frac{1}{x+h}) - (3x + \frac{1}{x})}{h} \] Simplifying this expression gives us: \[ f'(x) = \lim_{h \to 0} \frac{3h + \frac{1}{x+h} - \frac{1}{x}}{h} \] The terms \( 3x \) cancel out, leaving: \[ f'(x) = \lim_{h \to 0} \frac{3h + \left( \frac{1}{x+h} - \frac{1}{x} \right)}{h} \] Now, focusing on the \( \frac{1}{x+h} - \frac{1}{x} \) term, we can find a common denominator: \[ \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x + h)}{x(x+h)} = \frac{-h}{x(x+h)} \] Substituting this back into the limit, we have: \[ f'(x) = \lim_{h \to 0} \frac{3h - \frac{h}{x(x+h)}}{h} \] We can simplify this expression: \[ = \lim_{h \to 0} \left( 3 - \frac{1}{x(x+h)} \right) \] Now, as \( h \to 0 \), \( x + h \to x \), so: \[ f'(x) = 3 - \frac{1}{x^2} \] Thus, the derivative of the function \( f(x) = 3x + \frac{1}{x} \) is: \[ f'(x) = 3 - \frac{1}{x^2} \]