We have: #y = ln(sqrt(x - 1) - sqrt(x + 1))#
#Rightarrow y = ln((x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2)))#
This function can be differentiated using the "chain rule".
Let #u = (x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2))# and #v = ln(u)#:
#Rightarrow y' = u' cdot v'#
#Rightarrow y' = (frac(1)(2) (x - 1)^(- frac(1)(2)) - frac(1)(2) (x + 1)^(- frac(1)(2))) cdot frac(1)(u)#
#Rightarrow y' = frac(1)(u) cdot (frac(1)(2 sqrt(x - 1)) - frac(1)(2 sqrt(x + 1)))#
#Rightarrow y' = frac(1)(u) cdot (frac(1)(2) cdot (frac(1)(sqrt(x - 1)) - frac(1)(sqrt(x + 1))))#
#Rightarrow y' = frac(1)(2 u) (frac(sqrt(x + 1))(sqrt(x + 1) sqrt(x - 1)) - frac(sqrt(x - 1))(sqrt(x + 1) sqrt(x - 1)))#
#Rightarrow y' = frac(1)(2 u) (frac(sqrt(x + 1) - sqrt(x - 1))(sqrt(x + 1) sqrt(x - 1)))#
Let's replace #u# with #(x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2))#:
#Rightarrow y' = frac(1)(2 ((x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2)))) (frac(sqrt(x + 1) - sqrt(x - 1))(sqrt(x + 1) sqrt(x - 1)))#
#Rightarrow y' = frac(sqrt(x + 1) - sqrt(x - 1))(2 cdot (sqrt(x - 1) - sqrt(x + 1)) cdot sqrt(x + 1) sqrt(x - 1)))#
#Rightarrow y' = frac((sqrt(x + 1) - sqrt(x - 1)))(2 cdot - 1 cdot (sqrt(x + 1) - sqrt(x - 1)) cdot sqrt(x + 1) sqrt(x - 1))#
#Rightarrow y' = - frac(1)(2 cdot sqrt(x + 1) sqrt(x - 1))#
#Rightarrow y' = - frac(1)(2 sqrt((x + 1)(x - 1)))#
The argument of the radical can be factorised as a difference of two squares:
#Rightarrow y' = - frac(1)(2 sqrt(x^(2) - 1))# #\ # (shown.)