f(x)=(2x-1)(x+3)-x,x0=11/3
f`(x)=((2x–1)(x+3)–x)`=(2x-1)`*(x+3)-(2x-1)*(x+3)`-(x)`=2*(x+3)-(2x-1)*1-1=2x+6-2x+1-1=6 f`(x)=6 x_(o)=1(1/3) f`(x_(o))=[b]6[/b]