Y(x)=1 y`(x)=(1)`=0
y(x)=x y`(x)=(x)`=1
y=2x y`(x)=(2x)`=2
y=x² y`(x)=(x²)`=2x
y=3x³+3 y`(x)=(3x³+3)`=3*3x²+0=9x²
y=(2x³-3)(3x²-2) y`(x)=((2x³-3)(3x²-2))`=(2x³-3)`(3x²-2)+(2x³-3)(3x²-2)`=
=6x²(3x²-2)+(2x³-3)*6x=6x(x(3x²-2)+2x³-3)=
=6x(3x³-2x+2x³-3)=6x(5x³-2x-3)