Question

Решите 2 варианта во вложении,по производным

Answer 1

1) y'=(-x^3+0,5x^2-x+1)'= -3x^2+x-1

2) y'=(-3cosx)' * (x^2+2) + (-3cosx)*(x^2+2)'=3sinx(x^2+2)-3cosx*2x=3sinx(x^2+2)-6xcosx
3) y'= (x^-1/2)'=-1/2 * x^-3/2 = -1/(2x^(3/2))
4) y'=(1/sinx)'=(1'*sinx - 1*sinx')/sin^2(x) = (0-cosx)/sin^2(x) = -cosx/sin^2(x) = -ctgx/sinx
5) y'=x^4/(3-x)=(x^4'(3-x)-x^4(3-x)')/(3-x)^2=(4x^3(3-x)+x^4)/(3-x)^2=(12x^3-4x^4+x^4)/(3-x)^2= (12x^3-3x^4)/(3-x)^2
6) y'= (x^2+ctgx)' = 2x - 1/sin^2(x)

1) y' = -2x^3 + x^2 -2
2) y'=(4sqrt(x)+3)'(1-1/x)+(4sqrt(x)+3)(1-1/x)'=(2/sqrt(x))(1-1/x)+(4sqrt(x)+3)(1/x^2)
3) y'=(-x^-3)'=3x^-4=3/x^4
4) y'=(3'*sinx - 3sinx')/sin^2(x)=(0-3cosx)/sin^2(x))=-3cosx/sin^2(x)=-3ctgx/sinx
5) y'=((x^2+4)'cosx - (x^2+4)cosx')/cos^2(x)=(2xcosx + sinx(x^2+4)/cos^2(x)
6) y'=x^2' *tgx + x^2 * tgx' = 2xtgx + x^2/cos^2(x)