1.f'(x)=4x³-x²+5x-0,3
2.f'(x)=1/2√x+(2/3)·x^(2/3-1)+0,4·x^(-1/4)'+0,4·x^(-1/4)'=
=1/2√x+(2/3)·x^(-1/3)-(1/4)·0,4x^(-5/4)=1/2√x+2/3∛x-1/10x^(5/4)
3.f'(x)=((1-x³)/(1+x)³)'= ((1-x³)'(1+x)³-
(1-x³)(1+x)³)')/(1-x³)²=
=((-3x²(1+x³)-3x²(1-x³))/(1-x³)²=(-3x²-3x⁵-3x²+3x⁵)/(1-x³)²=-6x²/(1-x³)²