Đáp án:
\[f'\left( x \right) = \frac{{ - x + 100}}{{2\sqrt x {{\left( {x + 100} \right)}^2}}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = \frac{{\sqrt x }}{{x + 100}}\\
\Rightarrow f'\left( x \right) = \frac{{\left( {\sqrt x } \right)'.\left( {x + 100} \right) - \left( {x + 100} \right)'.\sqrt x }}{{{{\left( {x + 100} \right)}^2}}}\\
= \frac{{\frac{1}{{2\sqrt x }}.\left( {x + 100} \right) - 1.\sqrt x }}{{{{\left( {x + 100} \right)}^2}}}\\
= \frac{{\left( {x + 100} \right) - \sqrt x .2\sqrt x }}{{2\sqrt x .{{\left( {x + 100} \right)}^2}}}\\
= \frac{{x + 100 - 2x}}{{2\sqrt x {{\left( {x + 100} \right)}^2}}}\\
= \frac{{ - x + 100}}{{2\sqrt x {{\left( {x + 100} \right)}^2}}}
\end{array}\)
