Как-то так:
)f'(x)=((x^3/6)-0,5x^2-3x+2)'=1/6*3x^2-0,5*2x-3+0=1/2x^2-x-3=(x^2/2)-x-3
2)g'(x)=((3+2x)/(x-5))'=((3+2x)'(x-5)-(3+2x)(x-5)')/((x-5)^2)=(2(x-5)-1(3+2x))/((x-5)^2)=
=(2x-10-3-2x))/((x-5)^2)=(-13)/((x-5)^2)
3)f'(x)=(x*корень(x))'=(x*x^(1/2))'=(x^(2/2)*x^(1/2))'=(x^(2/2+1/2))'=(x^(3/2))'=
=3/2*(x^(3/2-2/2)=3/2*(x^(1/2))=3/2*корень(x)
4)y'(x)=(tgx-2ctgx+5)'=(1/cos^2(x))-2(-1/sin^2(x))+0=(1/cos^2(x))+(2/sin^2(x))=
=1+tg^2(x)+2(1+ctg(x))=1+tg^2(x)+2+2ctg(x))=tg^2(x)+ctg^2(x)+2