1)
[tex]f'(x)=6x^2+4[/tex]
2)
[tex]f'(x)=-x^3+x^2-x[/tex]
3)
[tex]f'(x)=(1-3x^2)'(x^2+x)+(1-3x^2)(x^2+x)'=-6x(x^2+x)+(1-3x^2)(2x+1)=\\=-6x^3-6x^2+2x+1-6x^3-3x^2=-12x^3-9x^2+2x+1[/tex]
4)
[tex]f'(x)=[(x-2)^2]'(1-x^2)+(x-2)^2(1-x^2)'=\\=2(x-2)(1-x^2)+(x^2-4x+4)*(-2x)=2(x-x^3-2+2x^2)-2x^3+8x^2-8x=\\=2x-2x^3-4+4x^2-2x^3+8x^2-8x=-4x^3+12x^2-6x-4[/tex]
5)
[tex]f'(x)=\frac{(2x^2-x+1)'(3x-x^2)-(2x^2-x+1)(3x-x^2)'}{(3x-x^2)^2}=\\=\frac{(4x-1)(3x-x^2)-(2x^2-x+1)(3-2x)}{(3x-x^2)^2}=\frac{12x^2-4x^3-3x+x^2-6x^2+4x^3+3x-2x^2-3+2x}{(3x-x^2)^2}=\\=\frac{5x^2+2x-3}{(3x-x^2)^2}[/tex]
6)
[tex]f'(x)=\frac{[(x-2)^2]'(1-x)-(x-2)^2(1-x)'}{(1-x)^2}=\frac{2(x-2)(1-x)-(x^2-4x+4)*(-1)}{(1-x)^2}=\\=\frac{2(x-x^2-2+2x)+(x^2-4x+4)}{(1-x)^2}=\frac{2x-2x^2-4+4x+x^2-4x+4}{(1-x)^2}=\frac{-x^2+2x}{(1-x)^2}[/tex]
7)
[tex]f'(x)=-\frac{1}{(1-x^3)^2}*(-3x^2)-\frac{1}{x^4}*2x=\frac{3x^2}{(1-x^3)^2}-\frac{2}{x^3}=\frac{3x^5}{x^3(1-x^3)^2}-\frac{2(1-x^3)^2}{x^3(1-x^3)^2}=\\=\frac{3x^5-2(1-x^3)^2}{x^3(1-x^3)^2}=\frac{3x^5-2(1-2x^3+x^6)}{x^3(1-x^3)^2}=\frac{3x^5-2+4x^3-2x^6}{x^3(1-x^3)^2}=\frac{-2x^6+3x^5+4x^3-2}{x^3(1-x^3)^2}[/tex]
8)
[tex]f'(x)=(\sqrt{x})'(1-2x^2)+\sqrt{x}(1-2x^2)'=\frac{1}{2\sqrt{x}}(1-2x^2)+\sqrt{x}(-4x)=\\=\frac{1-2x^2}{2\sqrt{x}}-4x\sqrt{x}=\frac{1-2x^2}{2\sqrt{x}}-\frac{8x^2}{2\sqrt{x}}=\frac{-10x^2+1}{2\sqrt{x}}[/tex]
9)
[tex]f'(x)=\frac{(\sin x)'(1+\cos x)-\sin x(1+\cos x)'}{(1+\cos x)^2}=\frac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x)^2}=\frac{\cos x+\cos^2 x+\sin^2 x}{(1+\cos x)^2}=\\=\frac{1+\cos x}{(1+\cos x)^2}=\frac{1}{1+\cos x}[/tex]
