Cho hàm số \[f\left( x \right) = {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^3}\]. Hàm số có đạo hàm f′(x) bằng:

\[f(x) = {(\sqrt x - \frac{1}{{\sqrt x }})^3} = {(\sqrt x )^3} - 3{(\sqrt x )^2}.\frac{1}{{\sqrt x }} + 3\sqrt x {(\frac{1}{{\sqrt x }})^2} - {(\frac{1}{{\sqrt x }})^3}\]

\[f(x) = {x^{\frac{3}{2}}} - 3\sqrt x + \frac{3}{{\sqrt x }} - \frac{1}{{{x^{\frac{3}{2}}}}}\]

\[f(x) = {x^{\frac{3}{2}}} - 3\sqrt x + 3{x^{ - \frac{1}{2}}} - {x^{ - \frac{3}{2}}}\]

\[f\prime (x) = \frac{3}{2}{x^{\frac{3}{2} - 1}} - \frac{3}{{2\sqrt x }} + 3.( - \frac{1}{2}){x^{ - \frac{1}{2} - 1}} + \frac{3}{2}{x^{ - \frac{3}{2} - 1}}\]

\[f\prime (x) = \frac{3}{2}\sqrt x - \frac{3}{{2\sqrt x }} - \frac{3}{2}{x^{ - \frac{3}{2}}} + \frac{3}{2}{x^{ - \frac{5}{3}}}\]

\[f\prime (x) = \frac{3}{2}(\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }})\]

Đáp án cần chọn là: D