Answer
$$f'\left( x \right) = 4x - 3$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = 2{x^2} - 3x + 1 \cr
& {\text{Differentiate using the definition of the derivative}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{{\left( {x + h} \right)}^2} - 3\left( {x + h} \right) + 1 - \left( {2{x^2} - 3x + 1} \right)}}{h} \cr
& {\text{Simplifying}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {{x^2} + 2xh + {h^2}} \right) - 3\left( {x + h} \right) + 1 - \left( {2{x^2} - 3x + 1} \right)}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^2} + 4xh + 2{h^2} - 3x - 3h + 1 - 2{x^2} + 3x - 1}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4xh + 2{h^2} - 3h}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {4x + 2h - 3} \right) \cr
& {\text{Evaluate the limit}} \cr
& f'\left( x \right) = 4x + 2\left( 0 \right) - 3 \cr
& f'\left( x \right) = 4x - 3 \cr} $$
