Найдите f' (x0), если

1) f (x) = 1 / (2x + 7) ^4 - (1-x) ^3;

X0 = - 3

2) f (x) = cos4x + ctgx;

X0 = п/2

3) f (x) = sqrt (корень из) (x^2 - 8x + 12)

X0 = 4

4) f (x) = xsin (x/3 + п/6)

X0 = п

1) f' = - 4 * (2x+7) ^-5 * 2 - 3 * (1-x) ^2 * (-1) = - 8 / (2x+7) ^5 + 3 * (1-x) ^2

f' (-3) = - 8 / (-6+7) ^5 + 3 * (1+3) ^2 = - 8+3*16 = - 8+48 = 40

2) f' = (-sin4x) * 4 - 1/sin^2 x = - (4*sin4x + 1/sin^2 x)

f' (пи/2) = - (4*sin 2 пи + 1 / sin^2 (пи/2)) = - (4*0 + 1/1) = - 1

3) f' = [ (x^2-8x+12) ^ (1/2) ]' = 1/2 * ((x^2-8x+12) ^ (-1/2)) * (2x-8) = 2 * (x-4) / (2 * (x^2-8x+12) ^ (1/2)) =

= (x-4) / (x^2-8x+12) ^ (1/2) = (x-4) / корень (x^2-8x+12)

f' (4) = (4-4) / корень (16-32+12) = 0

4) f ' = sin (x/3+пи/6) + x*cos (x/3+пи/6) * 1/3 = sin (x/3+пи/6) + (x*cos (x/3+пи/6)) / 3

f' (пи) = sin (пи/3+пи/6) + (пи*cos (пи/3+пи/6)) / 3 = sin (пи/2) + (пи*cos * (пи/2)) / 3 = 1 + пи*0/3 = 1