Đáp án:
$\begin{array}{l}
1)\\
y = \sqrt {1 - x} \\
\Rightarrow y' = \dfrac{{\left( {1 - x} \right)'}}{{2\sqrt {1 - x} }} = \dfrac{{ - 1}}{{2\sqrt {1 - x} }} = - 2{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\\
\Rightarrow y'' = - 2.\left( { - \dfrac{1}{2}} \right).\left( {1 - x} \right)'.{\left( {1 - x} \right)^{ - \dfrac{1}{2} - 1}}\\
= - {\left( {1 - x} \right)^{\dfrac{{ - 3}}{2}}}\\
= \dfrac{{ - 1}}{{\left( {1 - x} \right)\sqrt {1 - x} }}\\
2)\\
y = \tan x + {\mathop{\rm cotx}\nolimits} \\
\Rightarrow y' = \dfrac{1}{{{{\cos }^2}x}} - \dfrac{1}{{{{\sin }^2}x}} = 0\\
\Rightarrow \dfrac{1}{{{{\cos }^2}x}} = \dfrac{1}{{{{\sin }^2}x}}\\
\Rightarrow {\cos ^2}x = {\sin ^2}x\\
\Rightarrow \left[ \begin{array}{l}
\cos x = \sin x\\
\cos x = - \sin x
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = k\pi \\
x + \dfrac{\pi }{4} = k\pi
\end{array} \right.\\
\Rightarrow x = \pm \dfrac{\pi }{4} + k\pi \\
3)y' = 2x + \dfrac{1}{{{x^2}}}\\
\Rightarrow y = {x^2} - \dfrac{1}{x}
\end{array}$
