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

Question

1279 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-18T18:52:14

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

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

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 $x$ $in$ $D_f$ :

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

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

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

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

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

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