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:

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