1) y' = 1/(1 + x) * 1/(2 * x^(1/2)) - 1/(2 * x^(1/2)) = 1/(2 * x^(1/2)) * (1/(1 + x) - 1) = 1/(2 * x^(1/2)) * (-x)/(1 + x) = -1/2 * x^(1/2)/(1 + x) = -x^(1/2)/(2 + 2x)
2) (e^(sin x * ln (x^2 + 1)))'= e^(sin x * ln (x^2 + 1)) * (sin x * ln (x^2 + 1))' = (x^2 + 1)^sin x * (sin x * ln (x^2 + 1))'
3) sh x^2 + ch y = 1
ch x^2 * 2x + sh y * y' = 0
Осталось выразить y'.