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

Question

849 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)#