Find the derivative of $f$ from the first principle,where $f$ is given by $f(x) = \frac{2x+3}{x-2}$.

The derivative of a function $f(x)$ by the first principle is given by $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Given $f(x) = \frac{2x+3}{x-2}$,we have:
$f'(x) = \lim_{h \to 0} \frac{\frac{2(x+h)+3}{x+h-2} - \frac{2x+3}{x-2}}{h}$
Taking the common denominator in the numerator:
$f'(x) = \lim_{h \to 0} \frac{(2x+2h+3)(x-2) - (2x+3)(x+h-2)}{h(x-2)(x+h-2)}$
Expanding the terms:
$f'(x) = \lim_{h \to 0} \frac{(2x^2 - 4x + 2xh - 4h + 3x - 6) - (2x^2 + 2xh - 4x + 3x + 3h - 6)}{h(x-2)(x+h-2)}$
Simplifying the numerator:
$f'(x) = \lim_{h \to 0} \frac{2x^2 - x + 2xh - 4h - 6 - 2x^2 - 2xh + x - 3h + 6}{h(x-2)(x+h-2)}$
$f'(x) = \lim_{h \to 0} \frac{-7h}{h(x-2)(x+h-2)}$
Canceling $h$ and taking the limit as $h \to 0$:
$f'(x) = \frac{-7}{(x-2)(x-2)} = -\frac{7}{(x-2)^2}$.