Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
y = \sqrt {\frac{{{x^3}}}{{x - 1}}} = {\left( {\frac{{{x^3}}}{{x - 1}}} \right)^{\frac{1}{2}}}\\
\Rightarrow y' = \frac{1}{2}.\left( {\frac{{{x^3}}}{{x - 1}}} \right)'.{\left( {\frac{{{x^3}}}{{x - 1}}} \right)^{ - \frac{1}{2}}}\\
= \frac{1}{2}.\frac{{\left( {{x^3}} \right)'.\left( {x - 1} \right) - \left( {x - 1} \right)'.{x^3}}}{{{{\left( {x - 1} \right)}^2}}}.{\left( {\frac{{{x^3}}}{{x - 1}}} \right)^{ - \frac{1}{2}}}\\
= \frac{1}{2}.\frac{{3{x^2}\left( {x - 1} \right) - {x^3}}}{{{{\left( {x - 1} \right)}^2}}}.{\left( {\frac{{{x^3}}}{{x - 1}}} \right)^{\frac{{ - 1}}{2}}}\\
= \frac{1}{2}.\frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}.{\left( {\frac{{{x^3}}}{{x - 1}}} \right)^{ - \frac{1}{2}}}\\
h,\\
y = \frac{{4x + 1}}{{\sqrt {{x^2} + 2} }}\\
\Rightarrow y' = \frac{{\left( {4x + 1} \right)'.\sqrt {{x^2} + 2} - \left( {\sqrt {{x^2} + 2} } \right)'.\left( {4x + 1} \right)}}{{{x^2} + 2}}\\
= \frac{{4.\sqrt {{x^2} + 2} - \frac{{\left( {{x^2} + 2} \right)'}}{{2\sqrt {{x^2} + 2} }}.\left( {4x + 1} \right)}}{{{x^2} + 2}}\\
= \frac{{4\sqrt {{x^2} + 2} - \frac{x}{{\sqrt {{x^2} + 2} }}\left( {4x + 1} \right)}}{{{x^2} + 2}}\\
= \frac{{4\left( {{x^2} + 2} \right) - x\left( {4x + 1} \right)}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\\
= \frac{{4{x^2} + 8 - 4{x^2} - x}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\\
= \frac{{ - x + 8}}{{\left( {{x^2} + 2} \right)\sqrt {{x^2} + 2} }}\\
i,\\
y = \frac{{\sqrt {4 + {x^2}} }}{x}\\
\Rightarrow y' = \frac{{\left( {\sqrt {4 + {x^2}} } \right)'.x - x'.\sqrt {4 + {x^2}} }}{{{x^2}}}\\
= \frac{{\frac{{\left( {4 + {x^2}} \right)'}}{{2\sqrt {4 + {x^2}} }}.x - 1.\sqrt {4 + {x^2}} }}{{{x^2}}}\\
= \frac{{\frac{x}{{\sqrt {4 + {x^2}} }}.x - \sqrt {4 + {x^2}} }}{{{x^2}}}\\
= \frac{{{x^2} - \left( {4 + {x^2}} \right)}}{{{x^2}\sqrt {4 + {x^2}} }} = \frac{{ - 4}}{{{x^2}.\sqrt {4 + {x^2}} }}
\end{array}\)
