Lời giải của GV Vungoi.vn
\(\begin{array}{l}f\left( x \right) = {\left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right)^3} = {\left( {\sqrt x } \right)^3} - 3{\left( {\sqrt x } \right)^2}.\dfrac{1}{{\sqrt x }} + 3\sqrt x {\left( {\dfrac{1}{{\sqrt x }}} \right)^2} - {\left( {\dfrac{1}{{\sqrt x }}} \right)^3}\\f\left( x \right) = {x^{\dfrac{3}{2}}} - 3\sqrt x + \dfrac{3}{{\sqrt x }} - \dfrac{1}{{{x^{\dfrac{3}{2}}}}}\\f\left( x \right) = {x^{\dfrac{3}{2}}} - 3\sqrt x + 3{x^{ - \dfrac{1}{2}}} - {x^{ - \dfrac{3}{2}}}\\f'\left( x \right) = \dfrac{3}{2}{x^{\dfrac{3}{2} - 1}} - \dfrac{3}{{2\sqrt x }} + 3.\left( { - \dfrac{1}{2}} \right){x^{ - \dfrac{1}{2} - 1}} + \dfrac{3}{2}{x^{ - \dfrac{3}{2} - 1}}\\f'\left( x \right) = \dfrac{3}{2}\sqrt x - \dfrac{3}{{2\sqrt x }} - \dfrac{3}{2}{x^{ - \dfrac{3}{2}}} + \dfrac{3}{2}{x^{ - \dfrac{5}{2}}}\\f'\left( x \right) = \dfrac{3}{2}\left( {\sqrt x - \dfrac{1}{{\sqrt x }} - \dfrac{1}{{x\sqrt x }} + \dfrac{1}{{{x^2}\sqrt x }}} \right)\end{array}\)
Đáp án cần chọn là: d
Cách tính đạo hàm bằng hàm hợp như sau:
$\begin{array}{l}
f'\left( x \right) = 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)'\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left[ {\left( {\sqrt x } \right)' - \left( {\frac{1}{{\sqrt x }}} \right)'} \right]\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left[ {\frac{1}{{2\sqrt x }} - \frac{{ - \left( {\sqrt x } \right)'}}{{{{\left( {\sqrt x } \right)}^2}}}} \right]\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left( {\frac{1}{{2\sqrt x }} + \frac{{\frac{1}{{2\sqrt x }}}}{x}} \right)\\
= 3{\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{2x\sqrt x }}} \right)\\
= \frac{3}{2}\left( {x - 2 + \frac{1}{x}} \right)\left( {\frac{1}{{\sqrt x }} + \frac{1}{{x\sqrt x }}} \right)\\
= \frac{3}{2}\left( {\frac{x}{{\sqrt x }} - \frac{2}{{\sqrt x }} + \frac{1}{{x\sqrt x }} + \frac{x}{{x\sqrt x }} - \frac{2}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\\
= \frac{3}{2}\left( {\sqrt x - \frac{2}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)\\
= \frac{3}{2}\left( {\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{x\sqrt x }} + \frac{1}{{{x^2}\sqrt x }}} \right)
\end{array}$
