(lny)`+(e^(sin(x+y))`=0
так как
(lny)=y`/y
и
(e^(sin(x+y)))`=e^(sin(x+y))*(sin(x+y))`=e^(sin(x+y))*(cos(x+y))*(x+y)`=
=e^(sin(x+y))*(cos(x+y))*(1+y`)
то
y`/y + e^(sin(x+y))*(cos(x+y))*(1+y`)=0
y`/y + e^(sin(x+y))*(cos(x+y))+y`*e^(sin(x+y))*(cos(x+y))=0
y`*(e^(sin(x+y))*(cos(x+y)+(1/y))=-e^(sin(x+y))*(cos(x+y))
y`=-e^(sin(x+y))*(cos(x+y))/[b]([/b]e^(sin(x+y))*(cos(x+y)+(1/y))[b])[/b] - о т в е т