Đáp án:
$\begin{array}{l}
y = \sqrt {x + \sqrt {x + \sqrt x } } \\
= {\left( {x + \sqrt {x + \sqrt x } } \right)^{\frac{1}{2}}}\\
= {\left( {x + {{\left( {x + \sqrt x } \right)}^{\frac{1}{2}}}} \right)^{\frac{1}{2}}}\\
\Rightarrow y' = \frac{1}{2}.\left( {x + {{\left( {x + \sqrt x } \right)}^{\frac{1}{2}}}} \right)'.{\left( {x + {{\left( {x + \sqrt x } \right)}^{\frac{1}{2}}}} \right)^{ - \frac{1}{2}}}\\
= \frac{1}{2}.\left( {1 + \frac{1}{2}\left( {x + \sqrt x } \right)'.{{\left( {x + \sqrt x } \right)}^{ - \frac{1}{2}}}} \right).\frac{1}{{\sqrt {x + \sqrt {x + \sqrt x } } }}\\
= \frac{1}{2}.\frac{1}{{\sqrt {x + \sqrt {x + \sqrt x } } }}.\left( {1 + \frac{1}{2}.\left( {1 + \frac{1}{{2\sqrt x }}} \right).\frac{1}{{\sqrt {x + \sqrt x } }}} \right)\\
= \frac{1}{2}.\frac{1}{{\sqrt {x + \sqrt {x + \sqrt x } } }}\left( {1 + \frac{1}{2}.\frac{1}{{\sqrt {x + \sqrt x } }} + \frac{1}{4}.\frac{1}{{\sqrt {{x^2} + x} }}} \right)
\end{array}$
