Verified answer

1) у'=(-3x)'cosx+(-3x)(cosx)'=-3cosx+3xsinx

2)[tex]y'=\frac{(3x-2x^{2} )'(5x-1)-(3x-2x^{2} )(5x-1)'}{(5x-1)^{2} } =\frac{(3-4x)(5x-1)-(3x-2x^{2} )*5}{(5x-1)^{2}} =\frac{-10x^{2} +4x-3}{(5x-1)^{2} }[/tex]

3)y'=(2sinx)'+(4x⁻³)'-(5x)'=2cosx-12x⁻⁴-5= 2cosx-12/x⁴-5

[tex]\displaystyle\bf\\1)\\\\y=-3xCosx\\\\\\y'=-3\cdot\Big[x'\cdot Cosx+x\cdot(Cosx)'\Big]=-3\cdot\Big(1\cdot Cosx+x\cdot(-Sinx)\Big)=\\\\\\=-3\cdot\Big(Cosx-xSinx\Big)=3\cdot\Big(xSinx-Cosx\Big)\\\\\\2)\\\\y=\frac{3x-2x^{2} }{5x-1} \\\\\\y'=\frac{(3x-2x^{2} )'\cdot(5x-1)-(3x-2x^{2} )\cdot(5x-1)'}{(5x-1)^{2} }= \\\\\\=\frac{(3-4x)\cdot(5x-1)-(3x-2x^{2} )\cdot5}{(5x-1)^{2} } =\frac{15x-3-20x^{2} +4x-15x+10x^{2} }{(5x-1)^{2} } =\\\\\\=\frac{-10x^{2} +4x-3}{(5x-1)^{2} }[/tex]

[tex]\displaystyle\bf\\3)\\\\y=2Sinx+\frac{4}{x^{3} } -5x=2Sinx+4x^{-3} -5x\\\\\\y'=2\cdot(Sinx)'+4\cdot(x^{-3} )'-5\cdot x'=2Cosx+4\cdot(-3x^{-4} )-5\cdot 1=\\\\\\=2Cosx-\frac{12}{x^{4} } -5[/tex]