Bài 3 trang 71 SGK Toán 11 Tập 2 Cánh diều - CD

a) Ta có: \(y' = \left( {4{x^3} - 3{x^2} + 2x + 10} \right)' = 12{x^2} - 6x + 2\)

b) Ta có: \(y' = \left( {\frac{{x + 1}}{{x - 1}}} \right)' = \frac{{1.(x - 1) - ( - 1)(x + 1)}}{{{{(x - 1)}^2}}} = \frac{{x - 1 + x + 1}}{{{{(x - 1)}^2}}} = \frac{{2x}}{{{{(x - 1)}^2}}}\)

c) Ta có: \(y' = \left( { - 2x\sqrt x } \right)' = - 2.\left( {1.\sqrt x + x.\frac{1}{{2\sqrt x }}} \right) = - 2.\left( {\frac{{x + x}}{{2\sqrt x }}} \right) = - \frac{{2x}}{{\sqrt x }} = - 2\sqrt x \)

d) Ta có: \(y' = \left( {3\sin x + 4\cos x - \tan x} \right)' = 3\cos x - 4\sin x + \frac{1}{{{{\cos }^2}x}}\)\( = \frac{{3{{\cos }^3}x - 4\sin x.{{\cos }^2}x + 1}}{{{{\cos }^2}x}}\)

e) Ta có: \(y' = \left( {{4^x} + 2{e^x}} \right)' = {4^x}.\ln 4 + 2{e^x}\)

f) Ta có: \(y' = \left( {x\ln x} \right)' = x'\ln x + x\left( {\ln x} \right)' = \ln x + x.\frac{1}{x} = \ln x + 1\)

-- Mod Toán 11 HỌC247