Giải thích các bước giải:
\(\begin{array}{l}
a,\\
f\left( x \right) = {x^3} + 2\\
\Rightarrow f'\left( x \right) = 3{x^2} \Rightarrow f'\left( 2 \right) = {3.2^2} = 12\\
b,\\
f\left( x \right) = {x^2} + 2x + 1\\
\Rightarrow f'\left( x \right) = 2x + 2 \Rightarrow f'\left( 1 \right) = 2.1 + 2 = 4\\
c,\\
f\left( x \right) = \frac{1}{x}\\
\Rightarrow f'\left( x \right) = - \frac{1}{{{x^2}}} \Rightarrow f'\left( 2 \right) = - \frac{1}{{{2^2}}} = - \frac{1}{4}\\
d,\\
f\left( x \right) = \frac{{x + 1}}{{x - 1}}\\
f'\left( x \right) = \frac{{\left( {x + 1} \right)'.\left( {x - 1} \right) - \left( {x - 1} \right)'\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{\left( {x - 1} \right) - \left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}\\
f'\left( 0 \right) = \frac{{ - 1}}{{{{\left( {0 - 1} \right)}^2}}} = - 1
\end{array}\)
