Find the derivative of f(x) = (2x + 3)/(x - 2) from the first principle.

f'(x) = lim_h→0 [f(x + h) - f(x)] / h= lim_h→0 [((2(x + h) + 3)/(x + h - 2) - (2x + 3)/(x - 2))/ h]

= lim_{h → 0} [((2x + 2h + 3)(x - 2) - (2x + 3)(x + h - 2) ) / (h (x + h - 2)(x - 2))]

= lim_{h → 0} [((2x + 3)(x - 2) + 2h(x - 2) - (2x + 3)(x - 2) - (2x + 3)h ) / (h (x + h - 2)(x - 2))]

= lim_{h → 0} [(2h(x - 2) - (2x + 3)h ) / (h (x + h - 2)(x - 2))]

= lim_{h → 0} [h(2(x - 2) - (2x + 3)) / (h (x + h - 2)(x - 2))]

= lim_{h → 0} [(2x - 4 - 2x - 3) / ( (x + h - 2)(x - 2) )]

= lim_{h → 0} [(-7) / ( (x + h - 2)(x - 2) )]

= -7 / ((x - 2)(x - 2))

= -7 / (x - 2)²