y=x2x3+7y = \frac{x^2}{x^3 + 7}y=x3+7x2 Показать решение Решение: y′=(x2x3+7)′=(x2)′⋅(x3+7)−x2⋅(x3+7)′(x3+7)2=2x(x3+7)−x2⋅3x2(x3+7)2=14x−x4(x3+7)2y' = \left(\frac{x^2}{x^3 + 7}\right)' = \frac{(x^2)'\cdot(x^3 + 7) - x^2\cdot(x^3 + 7)'}{(x^3 + 7)^2} = \frac{2x(x^3 + 7) -x^2\cdot3x^2}{(x^3 + 7)^2} = \frac{14x - x^4}{(x^3 + 7)^2}y′=(x3+7x2)′=(x3+7)2(x2)′⋅(x3+7)−x2⋅(x3+7)′=(x3+7)22x(x3+7)−x2⋅3x2=(x3+7)214x−x4