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