Помогите пожалуйста! 1вариант

F(x) = 7 ⇒ f `(x) = 0f(x) = 2x-3 ⇒  f `(x) = 2f(x) = 12-10x ⇒  f `(x) = -10f(x) = x² ⇒  f `(x) = 2xf(x) = 1 / x^11 ⇒ f `(x) = -11 * (1 / x^12)f(x) = 2x³+x²+1 ⇒  f `(x) = 6x²+2xf(x) = x² - (1/25)*x +3 ⇒  f `(x) = 2x - (1/25)f(x) = (x-4)^5 ⇒  f `(x) = 5*(x-4)^4f(x) = (7-4x)^4 ⇒ f `(x) = (-16)*(7-4x)^3f(x) = 4(x+1)^3 ⇒  f `(x) = 12*(x+1)^2f(x) = sin(x+2) ⇒  f `(x) = cos(x+2)f(x) = sin(-x) ⇒  f `(x) = -cos(-x)f(x) = sin(8x-7) ⇒  f `(x) = 8*cos(8x-7)f(x) = tg(3x-2) ⇒ f `(x) = 3 / cos²(3x-2)f(x) = √(2x-5)  ⇒  f `(x) = (1/2) * [ (2x-5)^(-1/2) ] * 2 = 1 / √(2x-5)