a,
Đặt $u=x^2+x-3; v=x-1$
$\to \begin{cases}u'=(x^2+x-3)'=2x+1\\v'=(x-1)'=1\end{cases}$
Ta có:
`y'=(u'v-uv')/(v^2)=((2x-1)(x-1)-(x^2+x-3).1)/((x-1)^2)`
`=(x^2-2x+2)/((x-1)^2)`
b,
`y=x^3/3-x^2+4x-sqrtx+1/x`
`->y'=(x^3/3-x^2+4x-sqrtx+1/x)'`
`=(x^3/3)'-(x^2)'+(4x)'-(sqrtx)'+(1/x)'`
`=1/3.(x^3)'-(x^2)'+4.(x)'-(sqrtx)'+(x^(-1))'`
`=1/3 .3.x^2-2x+4-1/(2sqrtx)-x^(-2)`
`=x^2-2x+4-1/(2sqrtx)-1/x^2`
