f'(1) если f(x)=(3-x^2)(x^2+6)

(1) f'(x) = ((3 - x²)(x² + 6))' = (3 - x²)' × (x² + 6) + (3 - x²) × (x² + 6)' = -2x(x² + 6) + 2x(3 - x²) = -2x³ - 12x + 6x - 2x³ = -4x³ - 6x(2) f'(1) = -4 × 1³ - 6 × 1 = -4 × 1 - 6 = -4 - 6 = -10