\(\begin{array}{l}a)\,\,y = \dfrac{1}{{1 - x}}\\ \Rightarrow y' = - \dfrac{{ - 1}}{{{{\left( {1 - x} \right)}^2}}} = \dfrac{1}{{{{\left( {1 - x} \right)}^2}}}\\ \Rightarrow y'' = - \dfrac{{2\left( {1 - x} \right)\left( { - 1} \right)}}{{{{\left( {1 - x} \right)}^4}}} = \dfrac{2}{{{{\left( {1 - x} \right)}^3}}}\\b)\,\,y = \dfrac{1}{{\sqrt {1 - x} }}\\ \Rightarrow y' = - \dfrac{{\dfrac{{ - 1}}{{2\sqrt {1 - x} }}}}{{1 - x}} = \dfrac{1}{{2{{\left( {\sqrt {1 - x} } \right)}^3}}}\\ \Rightarrow y'' = - \dfrac{{3\left( {1 - x} \right).\dfrac{{ - 1}}{{2\sqrt {1 - x} }}}}{{2{{\left( {\sqrt {1 - x} } \right)}^6}}} = \dfrac{3}{{4{{\left( {\sqrt {1 - x} } \right)}^5}}}\\c)\,\,y = \tan x\\ \Rightarrow y' = \dfrac{1}{{{{\cos }^2}x}}\\ \Rightarrow y'' = \dfrac{{2\cos x\sin x}}{{{{\cos }^4}x}} = \dfrac{{2\sin x}}{{{{\cos }^3}x}}\\d)\,\,y = {\cos ^2}x\\ \Rightarrow y' = - 2\cos x\sin x = - \sin 2x\\ \Rightarrow y'' = - 2\cos 2x\end{array}\)
