f) y=($x^{3}$-2)(1-$x^{2}$ )
$y^{'}$ =($x^{3}$-2)'(1-$x^{2}$ )+($x^{3}$-2)(1-$x^{2}$ )'
=3$x^{2}$ (1-$x^{2}$ )+($x^{3}$-2)(-2x)
=-5$x^{4}$+3$x^{2}$ +4x
k)y=$\frac{2x-1}{1-x}$
$y^{'}$ =$\frac{(2x-1)'*(1-x)-(2x-1)*(1-x)'}{(x-1)^{2} }$
=$\frac{2*(1-x)-(2x-1)*(-1)}{(x-1)^{2}}$
=$\frac{1}{(x-1)^{2}}$