उत तर:
# - (3 (x + 1)) / (2 (2x-1) ^ 2 sqrt ((x + 1) / (2x-1)) #
स पष ट करण:
# एफ (एक स) = य ^ एन #
#f '(x) = n xx (du) / dx xxu ^ (n-1) #
इस म मल म:# sqrt ((x + 1) / (2x-1)) = ((x + 1) / (2x-1)) ^ (1/2): #
# एन = 1/2, य = (एक स + 1) / (2x-1) #
# d / dx = 1/2 xx (1xx (2x-1) - 2xx (x + 1)) / (2x-1) ^ 2 xx ((x + 1) / (2x-1)) ^ (1 /) 2-1) #
# = 1 / 2xx (-3) / ((2x-1) ^ 2 xx ((x + 1) / (2x-1)) ^ (1 / 2-1) #
# = - (3 (x + 1)) / (2 (2x-1) ^ 2 ((x + 1) / (2x-1)) ^ (1/2) #
