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