\[\begin{array}{l}y' = \frac{{\left[ {x + \sqrt {x + \sqrt x } } \right]'}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\ = \frac{{1 + \left[ {\sqrt {x + \sqrt x } } \right]'}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\ = \frac{{1 + \frac{{\left[ {x + \sqrt x } \right]'}}{{2\sqrt {x + \sqrt x } }}}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\ = \frac{{1 + \frac{{1 + \frac{1}{{2\sqrt x }}}}{{2\sqrt {x + \sqrt x } }}}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\ = \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\left[ {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}\left[ {1 + \frac{1}{{2\sqrt x }}} \right]} \right]\end{array}\]
Đề bài
Tìm đạo hàm của hàm số sau:
\[y = \sqrt {x + \sqrt {x + \sqrt x } } .\]
Lời giải chi tiết
\[\begin{array}{l}
y' = \frac{{\left[ {x + \sqrt {x + \sqrt x } } \right]'}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\
= \frac{{1 + \left[ {\sqrt {x + \sqrt x } } \right]'}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\
= \frac{{1 + \frac{{\left[ {x + \sqrt x } \right]'}}{{2\sqrt {x + \sqrt x } }}}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\
= \frac{{1 + \frac{{1 + \frac{1}{{2\sqrt x }}}}{{2\sqrt {x + \sqrt x } }}}}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\\
= \frac{1}{{2\sqrt {x + \sqrt {x + \sqrt x } } }}\left[ {1 + \frac{1}{{2\sqrt {x + \sqrt x } }}\left[ {1 + \frac{1}{{2\sqrt x }}} \right]} \right]
\end{array}\]