When we are given a fraction say f(x)=(3-2x-x^2)/(x^2-1)f(x)=3−2x−x2x2−1. This comprises of two fractions - say one g(x)=3-2x-x^2g(x)=3−2x−x2 in numerator and the other h(x)=x^2-1h(x)=x2−1, in the denominator. Here we use quotient rule as described below.
Quotient rule states if f(x)=(g(x))/(h(x))f(x)=g(x)h(x)
then (df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2dfdx=dgdx×h(x)−dhdx×g(x)(h(x))2
Here g(x)=3-2x-x^2g(x)=3−2x−x2 and hence (dg)/(dx)=-2-2xdgdx=−2−2x and as h(x)=x^2-1h(x)=x2−1, we have (dh)/(dx)=2xdhdx=2x and hence
(df)/(dx)=((-2-2x)xx(x^2-1)-2x xx(3-2x-x^2))/(x^2-1)^2dfdx=(−2−2x)×(x2−1)−2x×(3−2x−x2)(x2−1)2
= (-2x^3-2x^2+2x+2-6x+4x^2+2x^3)/(x^2-1)^2−2x3−2x2+2x+2−6x+4x2+2x3(x2−1)2
= (2x^2-4x+2)/(x^2-1)^22x2−4x+2(x2−1)2
or (2(x-1)^2)/(x^2-1)^22(x−1)2(x2−1)2
= 2/(x+1)^22(x+1)2
Observe that (3-2x-x^2)/(x^2-1)=((1-x)(3+x))/((x+1)(x-1))=(-3-x)/(x+1)3−2x−x2x2−1=(1−x)(3+x)(x+1)(x−1)=−3−xx+1 and using quotient rule
(df)/(dx)=(-(x+1)-(-3-x))/(x+1)^2=2/(x+1)^2dfdx=−(x+1)−(−3−x)(x+1)2=2(x+1)2
