Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,\\
f'\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}}\\
b,\\
f'\left( {\dfrac{1}{2}} \right) = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{f\left( x \right) - f\left( {\dfrac{1}{2}} \right)}}{{x - \dfrac{1}{2}}} = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{\left( { - {x^2} + 2} \right) - \dfrac{7}{4}}}{{x - \dfrac{1}{2}}}\\
 = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{ - {x^2} + \dfrac{1}{4}}}{{x - \dfrac{1}{2}}} = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{ - \left( {x - \dfrac{1}{2}} \right)\left( {x + \dfrac{1}{2}} \right)}}{{x - \dfrac{1}{2}}}\\
 =  - \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \left( {x + \dfrac{1}{2}} \right) =  - \left( {\dfrac{1}{2} + \dfrac{1}{2}} \right) =  - 1\\
4,\\
a,\\
y = \dfrac{{\sin {x^2}}}{x}\\
 \Rightarrow y' = \dfrac{{\left( {\sin {x^2}} \right)'.x - x'.\sin {x^2}}}{{{x^2}}}\\
 = \dfrac{{\left( {{x^2}} \right)'.\cos {x^2} - 1.\sin {x^2}}}{{{x^2}}}\\
 = \dfrac{{2x.\cos {x^2} - \sin {x^2}}}{{{x^2}}}\\
b,\\
y = \left( {x - 2} \right)\sqrt {{x^2} + 1} \\
 \Rightarrow y' = \left( {x - 2} \right)'.\sqrt {{x^2} + 1}  + \left( {x - 2} \right).\sqrt {{x^2} + 1} '\\
 = 1.\sqrt {{x^2} + 1}  + \left( {x - 2} \right).\dfrac{{\left( {{x^2} + 1} \right)'}}{{2\sqrt {{x^2} + 1} }}\\
 = \sqrt {{x^2} + 1}  + \left( {x - 2} \right).\dfrac{{2x}}{{2\sqrt {{x^2} + 1} }}\\
 = \sqrt {{x^2} + 1}  + \dfrac{{x\left( {x - 2} \right)}}{{\sqrt {{x^2} + 1} }}\\
 = \dfrac{{{{\sqrt {{x^2} + 1} }^2} + x.\left( {x - 2} \right)}}{{\sqrt {{x^2} + 1} }}\\
 = \dfrac{{2{x^2} - 2x + 1}}{{\sqrt {{x^2} + 1} }}
\end{array}\)