1) y(t)=(t^2-1)/(t^2+1) 2) (t^3-3t+1)/(t^3+2) 3)f(x)=(2+1/x^2)/(3-5x^3) - вопрос №53501851212123313232123534 от Chelovekkkkkkk1701 21.05.2020 14:57

Question

Алгебра: 1) y(t)=(t^2-1)/(t^2+1) 2) (t^3-3t+1)/(t^3+2) 3)f(x)=(2+1/x^2)/(3-5x^3) 4) y(t)=√t (3^√t +2t)

Chelovekkkkkkk1701 · Answer

Решение1) y(t)=(t^2-1)/(t^2+1) y`(t) = [2t*(t² + 1) - 2t*(t² - 1)] / (t² + 1)² = (2t³ + 2t - 2t³ + 2t) /  (t² + 1)² == 4t /  (t² + 1)²2) y(t) = (t^3-3t+1)/(t^3+2) y`(t) = [(3t² - 3)*(t³ + 2) - 3t² * (t³ - 3t + 1)] / (t³ + 2)² = (3t⁵ + 6t²- 3t³ - 6 - 3t⁵ + 9t³ + 3t²) / (t³ + 2)² = (6t³ + 9t² - 6) / (t³ + 2)²3) f(x) = (2+1/x^2)/(3-5x^3)f`(x) = [(- 2/x³)*(3 - 5x³) - 15x² * (2 + 1/x²)] / (3 - 5x)²  4) y(t) = √t (3^√t +2t) y`(t) = (1/2√t)* (3^√t +2t)  + √t [3^√t * ln3 * (1/2√t)  + 2]...