In(x/x^2+1) diffrentiate?

1 Answer
Jan 18, 2018

f'(x)=-(x^2-1)/(x^3+x)

Explanation:

f(x)=ln(x/(x^2+1))

For f to be defined in RR we need x/(x^2+1)>0

x^2+1 is always >0 so we need x>0

As a result Domain is D_f=(0,+oo)

For xinD_f:

f'(x)=1/(x/(x^2+1))*(x/(x^2+1))'=((x^2+1)/x)(x/(x^2+1))' =

((x^2+1)/x)(((x)'(x^2+1)-x(x^2+1)')/(x^2+1)^2) =

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

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

-(cancel(x^2+1)/x)((x^2-1)/(x^2+1)^cancel(2)) =

-(x^2-1)/(x(x^2+1)) = -(x^2-1)/(x^3+x)