You're going to have to use thw whole arsenal for this one - chain rule, power rule, quotient rule.
Since you can write your function as
#y = ln(u)#, with #u = (x^2 +1)^5/sqrt(1-x)#
you can use the chain rule to differentiate it like this
#color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))#
#y^' = d/(du)ln(u) * d/dx((x^2 +1)^5/sqrt(1-x))#
Now focus on calculating #d/dx((x^2 +1)^5/sqrt(1-x)) = d/dx(a(x))#, which you can do by using the quotient rule, chain rule again, and power rule.
You will get
#d/dx(a(x)) = ([d/dx(x^2 + 1)^5] * sqrt(1-x) - (x^2 + 1)^5 * d/dx(sqrt(1-x)))/(sqrt(1-x))^2#
This can be further broken down into
#d/dx(a(x)) = ( [5(x^2 + 1)^4 * 2x] * sqrt(1-x) - (x^2 + 1)^5 * (-1/(2sqrt(1-x))))/(1-x)#
#d/dx(a(x)) = (10x * (x^2 + 1)^4 sqrt(1-x) + 1/2 * (x^2 + 1)^5 * 1/(sqrt(1-x)))/(1-x)#
#d/dx(a(x)) = ((x^2 + 1)^4 * (1-x)^(-1/2) * [10x(1-x) + 1/2(x^2 + 1)])/(1-x)#
#d/dx(a(x)) = ((x^2 + 1)^4 * (1-x)^(-1/2) * (20x - 20x^2 + x^2 + 1))/(2(1-x))#
#d/dx(a(x)) = ((x^2 + 1)^4 * (1-x)^(-1/2) * (-19x^2 + 20x + 1))/(2(1-x))#
#d/dx(a(x)) = 1/2 * (x^2 + 1)^4 * (1-x)^(-3/2) * (-19x^2 + 20x + 1)#
Plug this back into your target derivative to get
#y^' = 1/u * 1/2 * (x^2 + 1)^4 * (1-x)^(-3/2) * (-19x^2 + 20x + 1)#
#y^' = sqrt(1-x)/(x^2 + 1)^color(red)(cancel(color(black)(5))) * 1/2 * color(red)(cancel(color(black)((x^2 + 1)^4))) * (1-x)^(-3/2) * (-19x^2 + 20x + 1)#
#y^' = 1/2 * (1-x)^(-1) * (-19x^2 + 20x + 1)/(x^2 + 1)#
#y^' = color(green)((-19x^2 + 20x + 1)/(2(1-x)(x^2 + 1)))#