Đáp án:
$\begin{array}{l}
g)y = \dfrac{x}{{{x^2} + 1}}\\
\Rightarrow y' = \dfrac{{1.\left( {{x^2} + 1} \right) - x.2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \dfrac{{ - {x^2} + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}\\
i)y = x + \sqrt {2{x^2} + 1} \\
\Rightarrow y' = 1 + \dfrac{{4x}}{{2\sqrt {2{x^2} + 1} }} = 1 + \dfrac{{2x}}{{\sqrt {2{x^2} + 1} }}\\
k)y = \dfrac{1}{x} - \dfrac{1}{{x - 2}}\\
\Rightarrow y' = \dfrac{{ - 1}}{{{x^2}}} + \dfrac{1}{{{{\left( {x - 2} \right)}^2}}}\\
l)y = \dfrac{{x + 1}}{{3\sqrt x }}\\
\Rightarrow y' = \dfrac{{3\sqrt x - \left( {x + 1} \right).\dfrac{3}{{\sqrt x }}}}{{9x}}\\
= \dfrac{{3x - 3x - 3}}{{9x\sqrt x }} = \dfrac{{ - 1}}{{3x\sqrt x }}\\
2)\\
y = 2\sin x + \tan x - 3x\\
\Rightarrow y' = 2\cos x + \dfrac{1}{{{{\cos }^2}x}} - 3\\
Khi:x \in \left[ {0;\dfrac{\pi }{2}} \right)\\
\Rightarrow \cos x \in \left( {0;1} \right]\\
Dat:{\mathop{\rm cosx}\nolimits} = t \Rightarrow t \in \left( {0;1} \right]\\
Xet:y' = f\left( t \right) = \dfrac{1}{{{t^2}}} + 2t - 3\,khi:t \in \left( {0;1} \right]\\
\Rightarrow f'\left( t \right) = - \dfrac{2}{{{t^3}}} + 2 = 0 \Rightarrow t = 1\\
\Rightarrow \min \,f\left( t \right) = f\left( 1 \right) = 0\\
\Rightarrow y' \ge 0\,khi\,x \in \left[ {0;\dfrac{\pi }{2}} \right)
\end{array}$
Vậy hàm số đồng biến trên nửa đoạn đã cho.
