Question

срочно помогите функция​

Answer 1

Ответ:

$1) f(x) = 4 + 3x, f(x)' = 3\\ 2) f(x) = 4x - 8, f(x)' = 4\\3) f(x) = 2x + 2x^2 -4x^3, f(x)' = 2+4x-12x^2\\4) f(x) = 2x^3 - 4x^2 +6.8, f(x)' = 6x^2-8x\\5) f(x) = 0.5sin(x) -2cos(x) - 3, f(x)' = 0.5cos(x) + 2sin(x)$

$6) f(x) = 2x^{0.5} - 2x +3, f(x)' = \frac{1}{\sqrt{x}} - 2\\7) f(x) = (3x-2)cos(x), f(x)' = 3cos(x) - (3x - 2)sin(x)\\8) f(x) = (4+3x)^2, f(x)' = 24 + 18x\\9) f(x) = \frac{4+3x}{2-3x} , f(x)' = \frac{18}{(2-3x)^2} \\10) f(x) = tg(3x) + 4x - 1, f(x)' = \frac{3}{cos^2(3x)} + 4$

Объяснение:

$1) f(x) = 4 + 3x, f(x)' = (4)' + (3x)' = 0 + 3 = 3\\ 2) f(x) = 4x - 8, f(x)' = (4x)' + (-8)' = 4 + 0 = 4\\3) f(x) = 2x + 2x^2 -4x^3, f(x)' = (2x)' + (2x^2)' - (4x^3)' =\\= 2 + 2*2x^{(2-1)}- 3*4x^{(3-1)} = 2+4x-12x^2\\4) f(x) = 2x^3 - 4x^2 +6.8, f(x)' = (2x^3)' - (4x^2)' + (6.8)' =\\= 3*2x^{(3-1)} - 2*4x^{(2 - 1)} + 0 = 6x^2-8x\\5) f(x) = 0.5sin(x) -2cos(x) - 3, f(x)' = (0.5sin(x))' -(2cos(x))' - (3)' =\\= 0.5cos(x) - (-2sin(x)) + 0 = 0.5cos(x) + 2sin(x)$

$6) f(x) = 2x^{0.5} - 2x +3, f(x)' = (2x^{0.5})' - (2x)' + (3)' =\\= 0.5*2x^{(0.5 - 1)} - 2 + 0 = \frac{1}{\sqrt{x}} - 2\\7) f(x) = (3x-2)cos(x), f(x)' = (3x-2)'cos(x) + (3x-2)(cos(x))' =\\= ((3x)' - (2)')cos(x) + (3x - 2)(-sin(x)) = 3cos(x) - (3x - 2)sin(x)\\8) f(x) = (4+3x)^2, f(x)' = ((4+3x)^2)' =\\= 2*(4+3x)^{2 - 1}(4+3x)'=2(4+3x)((4)' + (3x)') =\\= 2(4+3x)(0 + 3) = 6(4+3x)= 24 + 18x\\9) f(x) = \frac{4+3x}{2-3x} , f(x)' = (\frac{4+3x}{2-3x})' = \frac{(4+3x)'(2-3x)-(4+3x)(2-3x)'}{(2-3x)^2} =$

$= \frac{3(2 - 3x)-(-3)(4+3x)}{(2-3x)^2}= \frac{6 - 9x +12 + 9x}{(2-3x)^2} = \frac{18}{(2-3x)^2} \\10) f(x) = tg(3x) + 4x - 1, f(x)' = (tg(3x))' + (4x)' - (1)' =\\= (3x)'(\frac{1}{cos^2(3x)} )+ 4 - 0 = \frac{3}{cos^2(3x)} + 4$