Question

Решите номер 48. Есть вложение.

Answer 1

$( dfrac{u}{x} )'= dfrac{u'v-uv'}{v^2} \\ y= dfrac{x-1}{x} \\ y'= dfrac{(x-1)'x-(x-1)x'}{x^2} =dfrac{x-(x-1)}{x^2} =dfrac{1}{x^2}$