Question

Y12.. of e^2x sin(2x-1) where y12 is differentiation of y 12 times.

Do answer Fast..........

Answer 1

y = e^2x.sin(2x -1) ----------(1)
differentiate wrt x
y1 = 2e^2x {sin(2x -1) + cos(2x -1) }
again, differentiate wrt x
y2 = 4e^2x{sin(2x -1) + cos(2x -1)} + 4e^2x{cos(2x -1) - sin(2x -1) }
y2 = 8e^2x .cos(2x -1)

similarly,
y4 = -64 e^2x.sin(2x -1)
from equation (1)
y4 = -64y --------(2)
y8 = -64y4
from equation (2)
y8 = -64 × -64y
again ,
y12 = (64)² × y4
from equation (2)
y12 = (64)² × -64y

hence, y12 = -(64)³ × y
y12 = -(64)³ × e^2x.sin(2x -1)