Bài tập 5.86 trang 213 SBT Toán 11

Ta có: 

\(\begin{array}{l}
y' = \frac{{\left( {\tan \sqrt x } \right)'.\sqrt x  - \tan \sqrt x .\left( {\sqrt x } \right)'}}{x}\\
 = \frac{{\frac{1}{{2\sqrt x .{{\cos }^2}\sqrt x }} - \tan \sqrt x .\frac{1}{{2\sqrt x }}}}{x}\\
 = \frac{{\frac{1}{{2{{\cos }^2}\sqrt x }} - \frac{{\sin \sqrt x }}{{2\sqrt x \cos \sqrt x }}}}{x}\\
 = \frac{{\frac{1}{{2{{\cos }^2}\sqrt x }} - \frac{{\sin \sqrt x }}{{2\sqrt x \cos \sqrt x }}}}{x}\\
 = \frac{{2\sqrt 2  - \sin \left( {2\sqrt x } \right)}}{{4x\sqrt x {{\cos }^2}\sqrt x }}
\end{array}\)

Vậy \(dy = \frac{{2\sqrt x  - \sin \left( {2\sqrt x } \right)}}{{4x\sqrt x {{\cos }^2}\sqrt x }}dx\)

-- Mod Toán 11 HỌC247