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f '(x) = límite para h→0 [f(x + h) - f(x)] / H
1) f(x + h) = (x + h)² + 2 (x + h) = x² + 2 h x + 2 x + h² + 2 h
2) f(x + h) - f(x) = 2 h x + h² + 2 h = (2 x + h + 2) h
3) [f(x + h) - f(x)] / h = 2 x + h + 2
Si h → 0 nos queda f '(x) = 2 x + 2
Saludos Herminio