Đáp án:
\(\dfrac{{15}}{2}x\sqrt x + \dfrac{1}{{2x\sqrt x }} + 12x + \dfrac{2}{{{x^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = \left( {\dfrac{1}{{2\sqrt x }}} \right)\left( {3{x^2} - \dfrac{1}{x}} \right) + \left( {6x + \dfrac{1}{{{x^2}}}} \right)\left( {\sqrt x + 2} \right)\\
= \dfrac{{3x\sqrt x }}{2} - \dfrac{1}{{2x\sqrt x }} + 6x\sqrt x + 12x + \dfrac{1}{{x\sqrt x }} + \dfrac{2}{{{x^2}}}\\
= \dfrac{{15}}{2}x\sqrt x + \dfrac{1}{{2x\sqrt x }} + 12x + \dfrac{2}{{{x^2}}}
\end{array}\)
