y=−xx2+1y=-\dfrac{x}{x^2+1}y=−x2+1x Показать решение Решение: y′=−(xx2+1)′=−(x)′⋅(x2+1)−x⋅(x2+1)′(x2+1)2=−1⋅(x2+1)−x⋅2x(x2+1)2=−−x2+1(x2+1)2=x2−1(x2+1)2y'=-\left(\dfrac{x}{x^2+1}\right)' = -\frac{(x)'\cdot(x^2 + 1) - x\cdot(x^2 + 1)'}{(x^2 + 1)^2} = -\frac{1 \cdot (x^2+1) - x \cdot 2x}{(x^2+1)^2} = -\frac{-x^2 + 1}{(x^2 + 1)^2} = \frac{x^2 - 1}{(x^2 + 1)^2}y′=−(x2+1x)′=−(x2+1)2(x)′⋅(x2+1)−x⋅(x2+1)′=−(x2+1)21⋅(x2+1)−x⋅2x=−(x2+1)2−x2+1=(x2+1)2x2−1