1) tg2x= -^3 2) f(x)=x-2/x+3 Найти f '(x), f '(-2)
1) tg2x=-arctg(3)+Пn tgx=-arctg(3/2)+П/2n,n есть Z 2)f'(x)=(x-2)'(x+3)-(x-2)(x+3)'/(x+3)^2 (x+3)-(x-2)/(x+3)^2=x+3-x+2/(x+3)^2=5/(x+3)^2 f'(-2)=5/(-2+3)^2=5/1=5