Найти произведение в точке х0

1) f (x) = 5x^3-4x^2. x0=2

2) f (x) = 2sinx+cosx-ctgx. x0=п/6

3) f (x) = 3 (2x-1) ^51. x0=2

4) f (x) = корень из числа {2x^2+1}. x0=7

5) f (x) = sinx+cosx/sinx-cosx. x0=п/2

6 f (x) = 4cos^2*2x. x0=п/6

(^-степень пример 2x^2) (два х в квадрате)

1) f' (x) = (5x³-4x²) '=15x²-8x

f' (2) = 15·4-8·2=44

2) f' (x) = (2sinx+cosx-ctgx) '=2 (sinx) ' + (cosx) ' - (ctgx) '=

= 2cox-sinx + (1/sin²x)

f' (π/6) = 2·cos (π/6) - sin (π/6) + (1/sin² (π/6)) = (2√3/2) - (1/2) + (1 / (1/4)) = √3-0,5+4=3,5+√3

3) f' (x) = (3 (2x-1) ⁵¹) '=3· (2x-1) ⁵⁰· (2x-1) '=6· (2x-1) ⁵⁰

f' (2) = 6· (2·2-1) ⁵⁰=6·3⁵⁰

4) f'' (x) = (√ (2x²+1)) ' = (1/2√ (2 х²+1)) · (2 х²+1) '=4x/2√ (2 х²+1) = 2 х/√ (2 х²+1)

f' (7) = 14/√99

5) f' (x) = (sinx+cosx/sinx-cosx) ' = (sinx+cox) '· (sinx-cosx) - (sinx+cosx) · (sinx-cosx) ' / (sinx-cosx) ²=

= (cosx-sinx) (sinx-cosx) - (sinx+cosx) (cosx+sinx) / (sinx-cosx) ²=

=-4 (sin²x+cos²x) / (sinx-cosx) ²=-4 / (sinx-cosx) ²

f (п/2) = - 4 / (1-0) ²=-4

6) f' (x) = (4cos²2x) '=8cos2x· (cos2x) '=8cos2x· (-sin2x) · (2x) '=-8sin4x

f' (π/6) = - 8sin (2π/3) = - 8sin (π/3) = - 4√3