Đáp án:
\[y' = \mathop {\lim }\limits_{x \to 2} \frac{{ - 3x + 1}}{{2{{\left( {x + 1} \right)}^2}.\sqrt {3x + 1} }}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \frac{{\sqrt {3x + 1} }}{{x + 1}}\,\,\,\,\,\left( {x \ne - 1} \right)\\
\Rightarrow y' = \frac{{\left( {\sqrt {3x + 1} } \right)'.\left( {x + 1} \right) - \left( {x + 1} \right)'.\sqrt {3x + 1} }}{{{{\left( {x + 1} \right)}^2}}}\\
= \frac{{\frac{{\left( {3x + 1} \right)'}}{{2\sqrt {3x + 1} }}.\left( {x + 1} \right) - 1.\sqrt {3x + 1} }}{{{{\left( {x + 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{3}{{2\sqrt {3x + 1} }}\left( {x + 1} \right) - \sqrt {3x + 1} }}{{{{\left( {x + 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{3\left( {x + 1} \right) - 2{{\sqrt {3x + 1} }^2}}}{{2{{\left( {x + 1} \right)}^2}.\sqrt {3x + 1} }}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{3x + 3 - 6x - 2}}{{2{{\left( {x + 1} \right)}^2}.\sqrt {3x + 1} }}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{ - 3x + 1}}{{2{{\left( {x + 1} \right)}^2}.\sqrt {3x + 1} }}
\end{array}\)
