Đáp án:
\[y' = \frac{{ - 3}}{{2{x^2}\sqrt x }}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \frac{1}{{x\sqrt x }} = \frac{1}{{x.{x^{\frac{1}{2}}}}} = \frac{1}{{{x^{\frac{3}{2}}}}} = {x^{\frac{{ - 3}}{2}}}\\
\Rightarrow y' = \left( { - \frac{3}{2}} \right).{x^{ - \frac{3}{2} - 1}} = - \frac{3}{2}.{x^{ - \frac{5}{2}}} = - \frac{3}{2}.\frac{1}{{{x^{\frac{5}{2}}}}} = - \frac{3}{2}.\frac{1}{{{x^2}\sqrt x }} = \frac{{ - 3}}{{2{x^2}\sqrt x }}
\end{array}\)
