Tính đạo hàm các hàm số sau:

a) $y={\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^3$

b) y = 2^x + log₃(1 – 2x);

c) $y=\frac{1-2x}{x^2+1}$                 

d) y = sin2x + cos²3x.

a) $y^{\text{'}}={\left[{\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^3\right]}^{\text{'}}$

$=3{\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^2{\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^{\text{'}}$

$=3{\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^2\left(2x+\frac{2}{x^2}+\frac{2}{\sqrt{x}}\right)$

$=6{\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^2\left(x+\frac{1}{x^2}+\frac{1}{\sqrt{x}}\right)$

Vậy $y^{\text{'}}=6{\left(x^2-\frac{2}{x}+4\sqrt{x}\right)}^2\left(x+\frac{1}{x^2}+\frac{1}{\sqrt{x}}\right)$

b) y' = [2^x + log₃(1 – 2x)]' = $2^x\ln 2+\frac{{\left(1-2x\right)}^{\text{'}}}{\left(1-2x\right)\ln 3}$$=2^x\ln 2-\frac{2}{\left(1-2x\right)\ln 3}$

Vậy $y^{\text{'}}=2^x\ln 2-\frac{2}{\left(1-2x\right)\ln 3}$

c) $y^{\text{'}}={\left(\frac{1-2x}{x^2+1}\right)}^{\text{'}}$$=\frac{{\left(1-2x\right)}^{\text{'}}\left(x^2+1\right)-\left(1-2x\right){\left(x^2+1\right)}^{\text{'}}}{{\left(x^2+1\right)}^2}$

$=\frac{-2\left(x^2+1\right)-2x\left(1-2x\right)}{{\left(x^2+1\right)}^2}$$=\frac{-2x^2-2-2x+4x^2}{{\left(x^2+1\right)}^2}$$=\frac{2x^2-2x-2}{{\left(x^2+1\right)}^2}$

Vậy $y^{\text{'}}=\frac{2x^2-2x-2}{{\left(x^2+1\right)}^2}$

d) y' = (sin2x + cos²3x)' = cos2x×(2x)' + 2cos3x×(cos3x)'

= 2cos2x – 6cos3xsin3x = 2cos2x – 3sin6x.

Vậy y' = 2cos2x – 3sin6x.