(A) y′ = e−x(−2sin 2x) + cos 2x( −e−x).(C) y′ = (2 sec x)(sec x tan x).(E) + ln^3 x. The correct answer is 3 ln^2 x + ln^3 x.(B)(C)(D) Let y′ be ; then 3x^2 − 3y^2 y′ = 0; (A)(D) cos x + sin y · y′ = 0; (B) 6 x − 2(xy′ + y) + 10 yy′ = 0; y′(10y − 2x) = 2y − 6x.(A)