When we are given a fraction say #f(x)=(3-2x-x^2)/(x^2-1)#. This comprises of two fractions - say one #g(x)=3-2x-x^2# in numerator and the other #h(x)=x^2-1#, in the denominator. Here we use quotient rule as described below.

**Quotient rule** states if #f(x)=(g(x))/(h(x))#

then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

Here #g(x)=3-2x-x^2# and hence #(dg)/(dx)=-2-2x# and as #h(x)=x^2-1#, we have #(dh)/(dx)=2x# and hence

#(df)/(dx)=((-2-2x)xx(x^2-1)-2x xx(3-2x-x^2))/(x^2-1)^2#

= #(-2x^3-2x^2+2x+2-6x+4x^2+2x^3)/(x^2-1)^2#

= #(2x^2-4x+2)/(x^2-1)^2#

or #(2(x-1)^2)/(x^2-1)^2#

= #2/(x+1)^2#

Observe that #(3-2x-x^2)/(x^2-1)=((1-x)(3+x))/((x+1)(x-1))=(-3-x)/(x+1)# and using quotient rule

#(df)/(dx)=(-(x+1)-(-3-x))/(x+1)^2=2/(x+1)^2#