1) a) \(y = \tan x - 2{x^3} \Rightarrow y' = \frac{1}{{{{\cos }^2}x}} - 6{x^2}\)

b) \(y = x.\sin x + \sqrt {1 + {{\cos }^2}2x}  \Rightarrow y' = \sin x + x\cos x - \frac{{\sin 4x}}{{\sqrt {1 + {{\cos }^2}2x} }}\)

2) Ta có: \(y' = x - 3 \Rightarrow y'\left( { - 2} \right) =  - 5\)

\(y_0=8\)

Phương trình tiếp tuyến: \(y=5(x+2)+8 \Leftrightarrow y =  - 5x - 2\)

3) \(P\left( x \right) = \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right) \Rightarrow P'\left( x \right) = \left( {x - {x_2}} \right)\left( {x - {x_3}} \right) + \left( {x - {x_1}} \right)\left( {x - {x_3}} \right) + \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\)

Khi đó \(P\left( {{x_1}} \right) = \left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right);P'\left( {{x_2}} \right) = \left( {{x_2} - {x_1}} \right)\left( {{x_2} - {x_3}} \right);P'\left( {{x_3}} \right) = \left( {{x_3} - {x_1}} \right)\left( {{x_3} - {x_2}} \right)\)

Do đó \(\frac{1}{{P'\left( {{x_1}} \right)}} + \frac{1}{{P'\left( {{x_2}} \right)}} + \frac{1}{{P'\left( {{x_3}} \right)}} = \frac{{{x_3} - {x_2} + {x_1} - {x_3} + {x_2} - {x_1}}}{{\left( {{x_1} - {x_2}} \right)\left( {{x_2} - {x_3}} \right)\left( {{x_3} - {x_1}} \right)}} = 0\)