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