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

Giải thích thêm:

Cách tính đạo hàm bằng hàm hợp như sau:

$\begin{array}{l}
f'\left( x \right) = 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)'\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left[ {\left( {\sqrt x } \right)' - \left( {\frac{1}{{\sqrt x }}} \right)'} \right]\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left[ {\frac{1}{{2\sqrt x }} - \frac{{ - \left( {\sqrt x } \right)'}}{{{{\left( {\sqrt x } \right)}^2}}}} \right]\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left( {\frac{1}{{2\sqrt x }} + \frac{{\frac{1}{{2\sqrt x }}}}{x}} \right)\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2x\sqrt x }}} \right)\\
= \frac{3}{2}\left( {x - 2 + \frac{1}{x}} \right)\left( {\frac{1}{{\sqrt x }} + \frac{1}{{x\sqrt x }}} \right)\\
= \frac{3}{2}\left( {\frac{x}{{\sqrt x }} - \frac{2}{{\sqrt x }} + \frac{1}{{x\sqrt x }} + \frac{x}{{x\sqrt x }} - \frac{2}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\\
= \frac{3}{2}\left( {\sqrt x - \frac{2}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\\
= \frac{3}{2}\left( {\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)
\end{array}$