대답:
# = - 2csc2xcot2x #
설명:
방해
#f (x) = csc2x #
#f (x + Deltax) = csc2 (x + Deltax) #
#f (x + Deltax) -f (x) = csc2 (x + Deltax) -csc2x #
지금, = (csc2 (x + Deltax) -csc2x) / (Deltax) # (x)
# = 1 / (Deltax) ((csc2 (x + Deltax) -csc2x) / (Deltax)) #
# 1 / (Deltax) (1 / sin (2 (x + Deltax)) - 1 / sin (2x)) #
(sin2x-sin2 (x + Deltax)) / (sin (2 (x + Deltax)) sin2x)) # = 1 / (Deltax)
# sinC-sinD = 2cos ((C + D) / 2) sin ((C-D) / 2) #
의미하다
# C = 2x, D = 2 (x + Deltax) #
# (C + D) / 2 = (2x + 2 (x + Deltax)) / 2 #
# = (2x + 2x + 2Deltax) / 2 #
# = (4x + 2Deltax) / 2 #
# = 2 (2x + Deltax) / 2 #
# (C + D) / 2 = 2x + Deltax #
# (C-D) / 2 = (2x-2 (x + Deltax)) / 2 #
# = (2x-2x-2Deltax) / 2 #
# = (- 2Deltax) / 2 #
# (C-D) / 2 = -Deltax #
# sin2x-sin2 (x + Deltax) = 2cos (2x + Deltax) sin (-Deltax) #
(2x + Deltax) sin (-Deltax)) / (2x + Deltax) (sin (2 (x + Deltax)) sin2x) #
(1 / sin (2x)) ((cos (2x + Deltax)) / (sin (2 (x + Deltax))))) = (sin (Deltax)
sin (Deltax) 0) (sin (2x + Deltax)) / sinxlim (Deltaxto0)
#lim (Deltaxto0) (sin (Deltax) / (Deltax)) = 1 #
지금, # = - 2cscx (1) (cos2x) / sin (2x) #
# = - 2csc2xcot2x #
