Pytanie # 3dd7c

Odpowiedź:

# = - 2csc2xcot2x #

Wyjaśnienie:

Pozwolić

#f (x) = csc2x #

#f (x + Deltax) = csc2 (x + Deltax) #

#f (x + Deltax) -f (x) = csc2 (x + Deltax) -csc2x #

Teraz, #lim ((f (x + Deltax) -f (x)) / ((x + Deltax) -Deltax)) = (csc2 (x + Deltax) -csc2x) / (Deltax) #

# = 1 / (Deltax) ((csc2 (x + Deltax) -csc2x) / (Deltax)) #

# = 1 / (Deltax) (1 / sin (2 (x + Deltax)) - 1 / sin (2x)) #

# = 1 / (Deltax) ((sin2x-sin2 (x + Deltax)) / (sin (2 (x + Deltax)) sin2x)) #

# SinC-sinD = 2cos ((C + D) / 2) sin ((C-D) / 2) #

sugeruje

# 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) = 2 cosy (2x + Deltax) sin (-Deltax) #

#lim (Deltaxto0) ((f (x + Deltax) -f (x)) / ((x + Deltax) -Deltax)) = 1 / (Deltax) (2 cosy (2x + Deltax) sin (-Deltax)) / (sin (2 (x + Deltax)) sin2x) #

# = (2) (- sin (Deltax) / (Deltax)) (1 / sin (2x)) ((cos (2x + Deltax)) / (sin (2 (x + Deltax)))) #

# (- 2) / sinxlim (Deltaxto0) (sin (Deltax) / (Deltax)) lim (Deltaxto0) ((cos (2x + Deltax)) / (sin (2 (x + Deltax)))) #

#lim (Deltaxto0) (sin (Deltax) / (Deltax)) = 1 #

Teraz, # = - 2cscx (1) (cos2x) / sin (2x) #

# = - 2csc2xcot2x #