Đáp án:
$\begin{array}{l}
10)a)y = {\cos ^2}x\\
\Leftrightarrow y' = - 2\sin x.\cos x = - \sin 2x\\
b)y = {\tan ^3}x\\
\Leftrightarrow y' = 3.\dfrac{1}{{{{\cos }^2}x}}.{\tan ^2}x = \dfrac{{3.{{\sin }^2}x}}{{{{\cos }^4}x}}\\
d)y = {\cot ^2}\dfrac{x}{{1 - x}}\\
\Leftrightarrow y' = \left( {\dfrac{x}{{1 - x}}} \right)'.2.\dfrac{{ - 1}}{{{{\sin }^2}\dfrac{x}{{1 - x}}}}.\cot \dfrac{x}{{1 - x}}\\
= - 2.\dfrac{{1 - x + x}}{{{{\left( {1 - x} \right)}^2}}}.\dfrac{{\cos \dfrac{x}{{1 - x}}}}{{{{\sin }^3}\dfrac{x}{{1 - x}}}}\\
= \dfrac{{ - 2}}{{{{\left( {1 - x} \right)}^2}}}.\dfrac{{\cos \dfrac{x}{{1 - x}}}}{{{{\sin }^3}\dfrac{x}{{1 - x}}}}
\end{array}$
$\begin{array}{l}
c)y = \sin \sqrt {{x^2} + 1} \\
\Leftrightarrow y' = \left( {\sqrt {{x^2} + 1} } \right)'.cos\sqrt {{x^2} + 1} \\
= \dfrac{{2x}}{{2\sqrt {{x^2} + 1} }}.\cos \sqrt {{x^2} + 1} \\
= \dfrac{x}{{\sqrt {{x^2} + 1} }}.\cos \sqrt {{x^2} + 1}
\end{array}$
