Найти f ' (1):

1.)  f(x)=(x+1)^3 *(x+2)

2.)  f(x)=(x^2-1)/(x^2+1)

1)  f(x)=(x+1)³ *(x+2)

  f'(x)= 3(x+1)² *(x+2) + (x+1)³ = (x+1)² *(3(x+2) + (x+1))= (x+1)² *(3x+6 + x+1)=

  = (x+1)² *(4x+7)

  f'(1)= (1+1)² *(4 * 1+7) = (2)² *(4+7) = 4 * 11 = 44

2)  f(x)=(x²-1)/(x²+1)

  f'(x)= (2x(x²+1)-2x(x²-1))/(x²+1)² = (2x³+2x-2x³+2x)/(x²+1)² = 4x/(x²+1)² 

  f'(1)=  4 * 1/(1²+1)² = 4/(2)²= 4/4 = 1

1) f ' (x) = 3 * (x + 1)² * (x + 2) + (x + 1)³ * 1 = (x + 1)² * (3 * x + 6 + x + 1) =

(x + 1)² * (4 * x + 7)

f ' (1) = (1 + 1)² * (4 + 7) = 44

2) f (x) = 1 - 2 / (x² + 1)

    f ' (x) = 0 - 2 * (-2 * x) / (x² + 1)² = 4 * x / (x² + 1)²

    f ' (1) = 4 * 1 / 2² = 4 / 4 = 1