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<h1>Solución</h1>
<p>Siguiendo el mismo procedimiento que en el caso anterior, sin multiplicar por el conjugado:
$$\begin{aligned}
f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} {{\left( {{{\left( {x + h} \right)}^2} - 2\left( {x + h} \right)} \right) - \left( {{x^2} - 2x} \right)} \over h} &\text{Sustituyendo}\,\,\, f(x+h)\,\,\,\text{ y}\,\,\, f(x)\\
 &= \mathop {\lim }\limits_{h \to 0} {{{x^2} + 2xh + {h^2} - 2x - 2h - {x^2} + 2x} \over h}  &\text{Expandiendo}\,\,\,(x+h)^2-2(x+h)\\
 &= \mathop {\lim }\limits_{h \to 0} {{2xh - 2h + {h^2}} \over h} &\text{Simplificando}\\
& = \mathop {\lim }\limits_{h \to 0} {{h\left( {2x - 2 + h} \right)} \over h} &\text{Sacando}\,\,\, h \,\,\,\text{como factor cómún}\\
&=\mathop {\lim }\limits_{h \to 0} \left( {2x - 2 + h} \right) &\text{Cancelando h}\\
&=2x-2   &\text{Evaluando el límite}\\
\end{aligned}$$
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