1)
[tex]f'(x)=(2x^4+7x^3-5x^2+6x+1)'=8x^3+21x^2-10x+6[/tex]
2)
[tex]f'(x)=(\frac{-3}{5}x^5+\frac{1}{4}x^4-\frac{2}{3}x^3+\frac{1}{2}x^2-8x+2)'=-3x^4+x^3-2x^2+x-8[/tex]
3)
[tex]f'(x)=[(3x-5)(2x^2-3x+2)]'=(3x-5)'(2x^2-3x+2)+(3x-5)(2x^2-3x+2)'=3(2x^2-3x+2)+(3x-5)(4x-3)=6x^2-9x+6+12x^2-9x-20x+15=18x^2-38x+21[/tex]
4)
[tex]f'(x)=\left(\frac{x^2+2x}{x+3}\right)'=\frac{(x^2+2x)'(x+3)-(x^2+2x)(x+3)'}{(x+3)^2}=\frac{(2x+2)(x+3)-(x^2+2x)*1}{(x+3)^2}=\frac{2x^2+6x+2x+6-x^2-2x}{(x+3)^2}=\frac{x^2+6x+6}{(x+3)^2}[/tex]5)
[tex]f'(x)=[\sqrt{x}(2x^3+6x^2)]'=(\sqrt x)'(2x^3+6x^2)+\sqrt x(2x^3+6x^2)'=\frac{1}{2\sqrt x}(2x^3+6x^2)+\sqrt x(6x^2+12x)=\frac{\sqrt x}{2x}(2x^3+6x^2)+\sqrt x(6x^2+12x)=\sqrt x(x^2+3x)+\sqrt x(6x^2+12x)=\sqrt x(x^2+3x+6x^2+12x)=\sqrt x(7x^2+15x)[/tex]
6)
[tex]f'(x)=\left(\frac{\sqrt x}{4x+8}\right)'=\frac{(\sqrt x)'(4x+8)-\sqrt x(4x+8)'}{(4x+8)^2}=\frac{\frac{1}{2\sqrt x}(4x+8)-\sqrt x*4}{(4x+8)^2}=\frac{\frac{\sqrt x}{2 x}(4x+8)-\sqrt x*4}{(4x+8)^2}=\frac{\frac{\sqrt x}{ x}(2x+4)-\frac{\sqrt x}{x}*4x}{(4x+8)^2}=\frac{\frac{\sqrt x}{ x}(2x+4-4x)}{(4x+8)^2}=\frac{\sqrt x(-2x+4)}{x(4x+8)^2}=\frac{\sqrt x(4-2x)}{x(4x+8)^2}[/tex]
