This is equivalent to #f(z) = sqrt(z- 1)/sqrt(z + 1)#. We can easily differentiate this expression using the chain rule and the quotient rule.
Let #y = u^(1/2)# and #u = z - 1#.
#y' = 1/2u^(-1/2) = 1/(2u^(1/2))#
#u' = 1#
#dy/dx = 1 xx 1/(2u^(1/2)) = 1/(2(z - 1)^(1/2))#
The derivative of the denominator will be the same, except instead of #z - 1# in the denominator we will have #z + 1#.
Hence, #dy/dx = 1/(2(z + 1)^(1/2))#. We will now use the quotient rule to differentiate the entire expression.
#f'(z) = (1/(2(z - 1)^(1/2)) xx sqrt(z + 1) - 1/(2(z + 1)^(1/2)) xx sqrt(z - 1))/(sqrt(z + 1))^2#
#f'(z) = (sqrt(z + 1)/(2sqrt(z - 1)) - sqrt(z - 1)/(2sqrt(z + 1)))/(z + 1)#
This can be simplified.
#f'(z) = ((z + 1 - (z - 1))/(2sqrt(z - 1)sqrt(z+ 1)))/(z + 1)#
#f'(z) = 1/((z + 1)(sqrt(z - 1))(sqrt(z + 1))#
#f'(z) = 1/((z + 1)^(3/2)(z - 1)^(1/2))#
#f'(z) = (z + 1)^(-3/2)(z - 1)^(-1/2)#
Hopefully this helps!
