Нужна помощь с производными. Пожалуйста, срочно!

ЗАДАНИЕ 1. Найдите производную.y=3 => y'= 0y=2x => y'=2*1=2y=-x^2 => y'= -1*(x^2)' = -1*2*x^(2-1) = -2xy = x^5 => y'=5x^(5-1)=5x^4y=2x^3 - 3x => y'=(2x^3 - 3x)' = (2x^3)' - (3x)' = 2*(x^3)' - 3*(x)' = 2*(3x^(3-1)) - 3*1 = 6x^2 - 3y = 5x^2 + 3x - 4 => y' = (5x^2 + 3x - 4)' = 5*(x^2)' + 3*(x)' - (4)' = 5*2x + 3*1 - 0 = 10x+3y=-3x^8 + 3x^9-x+24 => y' = (-3x^8 + 3x^9-x+24)' = -3*(x^8)' + 3(x^9)' - (x)'+ (24)' == -3*8*x^7 + 3*9*x^8 - 1 + 0 = -24x^7 + 27x^8 - 1y = x^7 - 4x^5 + 2x - 6 => y' = (x^7 - 4x^5 + 2x - 6)' = (x^7)' - 4(x^5)' + 2(x)' - (6)' == 7x^6 - 4*5*x^(5-1) + 2*1 - 0 = 7x^6 - 20x^4 + 2f(x) = x => f '(x) = (x)' = 1f(x) = 3x-3 => f '(x) = (3x-3)' = 3(x)' - (3)' = 3*1-0 = 3f(x) = 4x^2 => f '(x) = (4x^2)' = 4(x^2)' = 4*2*x^(2-1) = 8xf(x) = -2x^3 + 4 => f '(x) = (-2x^3 + 4)' = -2(x^3)' + (4)' = -2*3*x^(3-1) + 0 = -6x^2f(x) = (-1/2)*x^2 + 5 => f '(x) = ((-1/2)*x^2 + 5)' = (-1/2)*(x^2)' + (5)' = (-1/2)*2*x^(2-1)+0 = -xf(x) = 4x^2 - 2x => f '(x) = (4x^2 - 2x)' = 4(x^2)' - 2(x)' = 8x - 2ЗАДАНИЕ 2. Найдите производные. y = 7x^(6/7) + 3x^(2/3) + 6 => y' = (7x^(6/7) + 3x^(2/3) + 6)' = 7(x^(6/7))' + 3(x^(2/3))' + (6)' = 7*(6/7)*x^(6/7 -1) + 3*(2/3)*x^(2/3 -1) + 0 = 6x^(-1/7) + 2x^(-1/3) = 6/x^7 + 21/x^3y = 9x^(2/9) + 14x^(2/7) + 5x^(2/5) + 6x^(1/3) + 4x + 2 =>=> y' = (9x^(2/9) + 14x^(2/7) + 5x^(2/5) + 6x^(1/3) + 4x + 2)' = = 9(x^(2/9))' + 14(x^(2/7))' + 5(x^(2/5))' + 6(x^(1/3))' + 4(x)' + (2)' = = 9*(2/9)*x^(2/9 -1) + 14*(2/7)*x^(2/7 - 1) + 5*(2/5)*x^(2/5 -1) + 6*(1/3)*x^(1/3 -1) + 4*1 + 0 = 2x^(-7/9) + 4x^(-5/7) + 2x^(-3/5) + 2*x^(-2/3) + 4 = = 2/x^(7/9) + 4/x^(5/7) + 2/x^(3/5) + 2/x^(2/3) + 4y = (x+1)/(x-2) => y' = ((x+1)/(x-2))' = ((x+1)' *(x-2) - (x+1)*(x-2)') / (x-2)^2 =(1*(x-2) - (x+1)*1) / (x-2)^2 = -3/(x-2)^2y = (5x-7)/(2x+9) => y' = ((5x-7)/(2x+9))' = ((5x-7)' *(2x+9) - (5x-7)*(2x+9)') /(2x+9)^2 = = (5*(2x+9) - (5x-7)*2) /(2x+9)^2 = (10x+45 - 10x-14) /(2x+9)^2 = 31/(2x+9)^2y= ((x-1)^2 / (2x+3)) => y'= (((x-1)^2)' *(2x+3) - ((x-1)^2) *(2x+3)') / (2x+3)^2 == (2*(x-1)*(2x+3) - 2*(x-1)^2) / (2x+3)^2 = 2*(x-1)*((2x+3) - (x-1)) / (2x+3)^2 == 2*(x-1)*(x+4) / (2x+3)^2y = (x^3 + 5x^2) / (3x-1) => y' = ((x^3 + 5x^2)' *(3x-1) - (x^3 + 5x^2) *(3x-1)' ) / (3x-1)^2 = = ((3x^2 + 10x)*(3x-1) - 3*(x^3 + 5x^2)) / (3x-1)^2 = = (9x^3 + 30x^2 - 3x^2 - 10x - 3x^3 - 15x^2) / (3x-1)^2 = (6x^3 + 12x^2 - 10x) / (3x-1)^2 = = 2x*(3x^2 + 6x - 5) / (3x-1)^2y = (3x^2 - 3x - 4) / (2x-1) => y' = ((3x^2 - 3x - 4)' *(2x-1) - (3x^2 - 3x - 4)*(2x-1)') / (2x-1)^2== ((6x-3)*(2x-1) - 2*(3x^2 - 3x - 4)) / (2x-1)^2 = (12x^2 - 12x + 3 - 6x^2 - 6x - 8) / (2x-1)^2== (6x^2 - 18x - 5) / (2x-1)^2y= (9x+5) / (x(x+1)) => y' = ((9x+5)' *(x(x+1)) - (9x+5) *(x(x+1))') / (x(x+1))^2 = = (9*(x(x+1)) - (9x+5) *(2x+1)) / (x(x+1))^2 = (-9x^2 -10x -5) / (x^2+x)^2 = = - (9x^2 +10x+5) / (x^2+x)^2

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