Đáp án:
\(a.y' = \dfrac{x}{{\sqrt {4 + {x^2}} }}.\cos \sqrt {4 + {x^2}} \)
Giải thích các bước giải:
\(\begin{array}{l}
a.y' = 2x.\dfrac{1}{{2\sqrt {4 + {x^2}} }}.\cos \sqrt {4 + {x^2}} \\
= \dfrac{x}{{\sqrt {4 + {x^2}} }}.\cos \sqrt {4 + {x^2}} \\
b.y' = - \dfrac{1}{{{x^2}}}.\dfrac{1}{{{{\cos }^2}\left( {x + \dfrac{1}{x}} \right)}}.\dfrac{1}{{2\sqrt {1 + \tan \left( {x + \dfrac{1}{x}} \right)} }}\\
d.y' = \dfrac{{2.2\cot 2x.\left( { - \dfrac{1}{{{{\sin }^2}2x}}} \right).\left( {x + 1} \right) - {{\cot }^2}2x}}{{{{\left( {x + 1} \right)}^2}}}\\
= \dfrac{{\dfrac{{ - 4\left( {x + 1} \right)\cos 2x}}{{{{\sin }^3}2x}} - \dfrac{{{{\cos }^2}2x.\sin 2x}}{{{{\sin }^3}2x}}}}{{{{\left( {x + 1} \right)}^2}}}\\
= \dfrac{{ - 4\left( {x + 1} \right)\cos 2x - {{\cos }^2}2x.\sin 2x}}{{{{\left( {x + 1} \right)}^2}{{\sin }^3}2x}}\\
e.y' = \dfrac{{2.\cos 2x.\sqrt {{x^2} + 1} - 2x.\dfrac{1}{{2\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}}\\
= \dfrac{{2\left( {{x^2} + 1} \right).\cos 2x - x}}{{\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}
\end{array}\)
