Tính đạo hàm của các hàm số sau:

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

b) $y=\frac{1-x^2}{x^2+1}$

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

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

$=\frac{\left(2x-1\right)\cdot \left(x+2\right)-\left(x^2-x+1\right)}{{\left(x+2\right)}^2}$$=\frac{2x^2+3x-2-x^2+x-1}{{\left(x+2\right)}^2}$$=\frac{x^2+4x-3}{{\left(x+2\right)}^2}$

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

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

 $=\frac{{\left(1-x^2\right)}^{\text{'}}\cdot \left(x^2+1\right)-\left(1-x^2\right)\cdot {\left(x^2+1\right)}^{\text{'}}}{{\left(x^2+1\right)}^2}$$=\frac{\left(-2x\right)\cdot \left(x^2+1\right)-\left(1-x^2\right)\cdot \left(2x\right)}{{\left(x^2+1\right)}^2}$$=\frac{-2x^3-2x-\left(2x-2x^3\right)}{{\left(x^2+1\right)}^2}$$=\frac{-4x}{{\left(x^2+1\right)}^2}$