Question

Найти производную: 1)y=x^3-2x^2+x+2 2)y=корень х*(2sinx+1) 3)y=0.25x^4+x^2-4 4)y=(x^2-5x)*(1-2x) 5)y=-x^3+0.5x^2-x+1 6)y=-3cosx 7)y=(3cosx-1)*(2sinx+2) 8)y=0.5x^4+1/3x^3-2x

Answer 1

$1) y= x^{3}-2 x^{2} +x+2 y'=3 x^{2}-4x+1 2) y= \sqrt{x} *(2sinx+1) y'=(\sqrt{x})'*(2sinx+1)+(2sinx+1)'* \sqrt{x} = 1/2\sqrt{x}*(2sinx+1)+2cosx*\sqrt{x}=(2sinx+1)/2 \sqrt{x}+2cosx*\sqrt{x} 3)y=0.25x^{4}+x^{2}-4 y'=x^{3}+2x 4)y=(x^{2}-5x)*(1-2x) y'=(x^{2}-5x)'*(1-2x)+(1-2x)'*(x^{2}-5x)= (2x-5)*(1-2x)-2*(x^{2}-5x)=2x-2x^{2}-5+10x-2x^{2}-10x= -4x^{2}+2x-5. 5)y=-x^{3}+0.5x^{2}-x+1 y'=-3x^{2}+x-1 6)y=-3cosx y'=3sinx 7)y=(3cosx-1)*(2sinx+2) y'=(3cosx-1)'*(2sinx+2)+(2sinx+2)'*(3cosx-1)= -3sinx*(2sinx+2)+2cosx*(3cosx-1)= -6sin^{2}x-6sinx+6cos^{2}x-2cosx= -6sinx(sinx+1)+cosx(6cosx-2) 8)y=0.5x^{4}+1/3x^{3}-2x y'=x^{3}+x^{2}-2$