Tính đạo hàm của các hàm số sau:a) \(y = \left( {1 - 2x} \right)\sqrt {1 + x - 2{x^2}} \)b) \(y = {\cos ^2}\left( {1 - 2{x^2}} \right)\

a) \(y' = \left( {1 - 2x} \right)'\sqrt {1 + x - 2{x^2}}  + \left( {1 - 2x} \right)\left( {\sqrt {1 + x - 2{x^2}} } \right)'\)

\(\begin{array}{l}
 =  - 2\sqrt {1 + x - 2{x^2}}  + \left( {1 - 2x} \right)\frac{{1 - 4x}}{{2\sqrt {1 + x - 2{x^2}} }}\\
 = \frac{{ - 4\left( {1 + x - 2{x^2}} \right) + \left( {1 - 2x} \right)\left( {1 - 4x} \right)}}{{2\sqrt {1 + x - 2{x^2}} }} = \frac{{16{x^2} - 10x - 3}}{{2\sqrt {1 + x - 2{x^2}} }}
\end{array}\)

b) \(y' = 2\cos \left( {1 - 2{x^2}} \right).\left( {{\rm{cos}}\left( {1 - 2{x^2}} \right)} \right)'\)

\(\begin{array}{l}
 =  - 2\cos \left( {1 - 2{x^2}} \right).\sin \left( {1 - 2{x^2}} \right).\left[ {\left( {1 - 2{x^2}} \right)'} \right]\\
 = 8x\cos \left( {1 - 2{x^2}} \right).\sin \left( {1 - 2{x^2}} \right)\\
 = 4x\sin \left( {2 - 4{x^2}} \right)
\end{array}\)