Calculus of a Single Variable, Chapter 2, 2.3, Section 2.3, Problem 25

You need to find derivative of the function using the quotient rule:
f'(x)= ((4 - 3x - x^2)'*(x^2-1) - (4 - 3x - x^2)*(x^2-1)')/((x^2-1)^2)
You need to use the quotient rule to differentiate ((x-1)/(x+1)):
f'(x)= ((-3 - 2x)*(x^2-1) - (4 - 3x - x^2)*(2x))/((x^2-1)^2)
f'(x)= (-3x^2 + 3 -2x^3 + 2x -8x+ 6x^2+ 2x^3)/((x^2-1)^2)
Reducing like terms yields:
f'(x)= (3x^2 - 6x + 3)/((x^2-1)^2)
You need to factor out (3):
f'(x)= 3(x^2 - 2x + 1)/((x^2-1)^2)
f'(x)= 3((x-1)^2)/((x-1)^2*(x+1)^2)
Simplify by (x-1)^2 :
f'(x)= 3/((x+1)^2)
Hence, evaluating the derivative of the function, yields f'(x)= 3/((x+1)^2).