Đề bài
Tìm vi phân của hàm số sau: \(y = {{\tan \sqrt x } \over {\sqrt x }}.\)
Sử dụng công thức \(dy = y'dx\).
Lời giải chi tiết
\(\begin{array}{l}
y'\\
= \dfrac{{\left( {\tan \sqrt x } \right)'.\sqrt x - \tan \sqrt x .\left( {\sqrt x } \right)'}}{{{{\left( {\sqrt x } \right)}^2}}}\\
= \dfrac{{\left( {\sqrt x } \right)'.\dfrac{1}{{{{\cos }^2}\sqrt x }}.\sqrt x - \tan \sqrt x .\dfrac{1}{{2\sqrt x }}}}{x}\\
= \dfrac{{\dfrac{1}{{2\sqrt x }}.\dfrac{{\sqrt x }}{{{{\cos }^2}\sqrt x }} - \dfrac{{\tan \sqrt x }}{{2\sqrt x }}}}{x}\\
= \dfrac{{\dfrac{1}{{2{{\cos }^2}\sqrt x }} - \dfrac{{\tan \sqrt x }}{{2\sqrt x }}}}{x}\\
= \dfrac{{\dfrac{{\sqrt x - \tan \sqrt x .{{\cos }^2}\sqrt x }}{{2\sqrt x {{\cos }^2}\sqrt x }}}}{x}\\
= \dfrac{{\sqrt x - \dfrac{{\sin \sqrt x }}{{\cos \sqrt x }}.{{\cos }^2}\sqrt x }}{{2x\sqrt x {{\cos }^2}\sqrt x }}\\
= \dfrac{{\sqrt x - \sin \sqrt x \cos \sqrt x }}{{2x\sqrt x {{\cos }^2}\sqrt x }}\\
= \dfrac{{2\sqrt x - 2\sin \sqrt x \cos \sqrt x }}{{4x\sqrt x {{\cos }^2}\sqrt x }}\\
= \dfrac{{2\sqrt x - \sin \left( {2\sqrt x } \right)}}{{4x\sqrt x {{\cos }^2}\sqrt x }}\\
\Rightarrow dy = y'dx\\
= \dfrac{{2\sqrt x - \sin \left( {2\sqrt x } \right)}}{{4x\sqrt x {{\cos }^2}\sqrt x }}dx
\end{array}\)
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