Đáp án:
$\begin{array}{l}
y = {x^2} - 2x + 12\\
\Rightarrow y' = 2x - 2\\
{y_0} = 12\\
\Rightarrow x_0^2 - 2{x_0} + 12 = 12\\
\Rightarrow \left[ \begin{array}{l}
{x_0} = 2\\
{x_0} = 0
\end{array} \right.\\
\Rightarrow PTTT:y = {y_0}'\left( {x - {x_0}} \right) + {y_0}\\
\Rightarrow \left[ \begin{array}{l}
y = \left( {2.2 - 2} \right).\left( {x - 2} \right) + 12\\
y = \left( {2.0 - 2} \right).\left( {x - 0} \right) + 12
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
y = 2x + 8\\
y = - 2x + 12
\end{array} \right.\\
2)a)y = 12{x^2} + 12x - 2020\\
\Rightarrow y' = 24x + 12\\
b)y = \left( {x + 12} \right)\left( {{x^2} - 12} \right)\\
\Rightarrow y' = {x^2} - 12 + \left( {x + 12} \right).2x\\
\Rightarrow y' = 3{x^2} + 24x - 12\\
c)y = {\left( {{x^{12}} - 12\sqrt x + 12x - 12} \right)^{12}}\\
\Rightarrow y' = 12.\left( {12{x^{11}} - \dfrac{6}{{\sqrt x }} + 12} \right).\\
{\left( {{x^{12}} - 12\sqrt x + 12x - 12} \right)^{11}}\\
d)y = \dfrac{{12x + 13}}{{12x + 23}} = 1 - \dfrac{{10}}{{12x + 23}}\\
\Rightarrow y' = - 10.\dfrac{{ - 12}}{{{{\left( {12x + 23} \right)}^2}}} = \dfrac{{120}}{{{{\left( {12x + 23} \right)}^2}}}\\
e)y = \sin \left( {\dfrac{{14x - 12}}{{6x + 12}}} \right)\\
\Rightarrow y' = \left( {\dfrac{{14x - 12}}{{6x + 12}}} \right)'.cos\left( {\dfrac{{14x - 12}}{{6x + 12}}} \right)\\
\Rightarrow y' = \dfrac{{240}}{{{{\left( {6x + 12} \right)}^2}}}.cos\left( {\dfrac{{14x - 12}}{{6x + 12}}} \right)\\
f)y = \sin \left( {\tan \,{x^{12}}} \right)\\
\Rightarrow y' = \left( {\tan \,{x^{12}}} \right)'.cos\left( {\tan \,{x^{12}}} \right)\\
\Rightarrow y' = 12.{x^{11}}.\dfrac{1}{{co{s^2}{x^{12}}}}.cos\left( {\tan \,{x^{12}}} \right)
\end{array}$
