Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = {\left( {\sqrt {{x^2} + x} + x} \right)^2}.\sqrt {{{\cot }^3}x} = {\left( {\sqrt {{x^2} + x} + x} \right)^2}.{\left( {\cot x} \right)^{\dfrac{3}{2}}}\\
\Rightarrow y' = \left[ {{{\left( {\sqrt {{x^2} + x} + x} \right)}^2}} \right]'.{\left( {\cot x} \right)^{\dfrac{3}{2}}} + {\left( {\sqrt {{x^2} + x} + x} \right)^2}.\left[ {{{\left( {\cot x} \right)}^{\dfrac{3}{2}}}} \right]'\\
= 2.\left( {\sqrt {{x^2} + x} + x} \right)'.\left( {\sqrt {{x^2} + x} + x} \right).{\left( {\cot x} \right)^{\dfrac{3}{2}}} + {\left( {\sqrt {{x^2} + x} + x} \right)^2}.\dfrac{3}{2}.\left( {\cot x} \right)'.{\left( {\cot x} \right)^{\dfrac{1}{2}}}\\
= 2.\left( {\dfrac{{\left( {{x^2} + x} \right)'}}{{2\sqrt {{x^2} + x} }} + 1} \right).\left( {\sqrt {{x^2} + x} + x} \right).\sqrt {{{\cot }^3}x} + {\left( {\sqrt {{x^2} + x} + x} \right)^2}.\dfrac{3}{2}.\dfrac{{ - 1}}{{{{\sin }^2}x}}.\sqrt {\cot x} \\
= 2.\left( {\dfrac{{2x + 1}}{{2\sqrt {{x^2} + x} }} + 1} \right)\left( {\sqrt {{x^2} + x} + x} \right).\sqrt {{{\cot }^3}x} - \dfrac{3}{2}{\left( {\sqrt {{x^2} + x} + x} \right)^2}.\dfrac{{\sqrt {\cot x} }}{{{{\sin }^2}x}}
\end{array}\)
