Ερώτηση # 3dd7c

Απάντηση:

# = - 2csc2xcot2x #

Εξήγηση:

Αφήνω

# f (x) = csc2x #

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

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

Τώρα, (x + Deltax) -f (x)) / ((x + Deltax) -Deltax)) = (csc2 (x + Deltax)

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

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

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

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

υποδηλώνει

# C = 2χ, D = 2 (χ + Deltax) #

# (C + D) / 2 = (2χ + 2 (χ + Deltax)) / 2 #

# = (2χ + 2χ + 2Deltax) / 2 #

# = (4x + 2Deltax) / 2 #

# = 2 (2χ + Deltax) / 2 #

# (C + D) / 2 = 2χ + Deltax #

# (C-D) / 2 = (2χ-2 (χ + Deltax)) / 2 #

# = (2x-2x-2Deltax) / 2 #

# = (- 2Deltax) / 2 #

# (C-D) / 2 = -Επεξεργασία #

# sin2x-sin2 (χ + Deltax) = 2cos (2χ + Deltax) sin (-Deltax) #

# del (Deltaxto0) ((f (x + Deltax) -f (x)) / ((χ + Deltax) -Deltax) (sin (2 (χ + Deltax)) sin2x) #

= (Sin (Deltax) / (Deltax)) (1 / sin (2x)) (cos (2x + Deltax)

((Deltax)) (sin (Deltax) / (Deltax)) lim (Deltaxto0) ((cos (2x + Deltax)

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

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

# = - 2csc2xcot2x #