Đáp án:

b. \(\left[ \begin{array}{l}
y = 2x - 1\\
y = 2x + 7
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a.f'\left( x \right) = \frac{{\cos 2x\left( {1 + \sin 2x} \right) + 2.x.\left( { - \sin 2x} \right)\left( {1 + \sin 2x} \right) - 2\cos 2x\left( {x.\cos 2x} \right)}}{{{{\left( {1 + \sin 2x} \right)}^2}}}\\
 = \frac{{\cos 2x + \cos 2x.\sin 2x - 2x\sin 2x - 2x.{{\sin }^2}2x - 2x.{{\cos }^2}2x}}{{{{\left( {1 + \sin 2x} \right)}^2}}}\\
 = \frac{{\cos 2x + \cos 2x.\sin 2x - 2x\sin 2x - 2x\left( {{{\sin }^2}2x + {{\cos }^2}2x} \right)}}{{{{\left( {1 + \sin 2x} \right)}^2}}}\\
 = \frac{{\cos 2x + \cos 2x.\sin 2x - 2x\sin 2x - 2x}}{{{{\left( {1 + \sin 2x} \right)}^2}}}\\
 = \frac{{\cos 2x\left( {1 + \sin 2x} \right) - 2x\left( {1 + \sin 2x} \right)}}{{{{\left( {1 + \sin 2x} \right)}^2}}}\\
 = \frac{{\left( {\cos 2x - 2x} \right)\left( {1 + \sin 2x} \right)}}{{{{\left( {1 + \sin 2x} \right)}^2}}}\\
 = \frac{{\cos 2x - 2x}}{{1 + \sin 2x}}\\
b.y' = \frac{{x + 1 - x + 1}}{{{{\left( {x + 1} \right)}^2}}} = \frac{2}{{{{\left( {x + 1} \right)}^2}}}\\
 \to y'\left( {{x_0}} \right) = \frac{2}{{{{\left( {{x_0} + 1} \right)}^2}}}\\
Do:tt//y = 2x + 1017\\
 \to y'\left( {{x_0}} \right) = k = 2\\
 \to \frac{2}{{{{\left( {{x_0} + 1} \right)}^2}}} = 2\\
 \to {\left( {{x_0} + 1} \right)^2} = 1\\
 \to \left| {{x_0} + 1} \right| = 1\\
 \to \left[ \begin{array}{l}
{x_0} + 1 = 1\\
{x_0} + 1 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x_0} = 0\\
{x_0} =  - 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{y_0} =  - 1\\
{y_0} = 3
\end{array} \right.\\
 \to PTTT:\left[ \begin{array}{l}
y = 2x - 1\\
y = 2\left( {x + 2} \right) + 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
y = 2x - 1\\
y = 2x + 7
\end{array} \right.
\end{array}\)