How to differentiate y= (ln x)/(1+ln x) ?

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Answer 1 · 2018-04-14T06:12:33

#y'(x)=1/(x(1+lnx)^2)#

#d/dx(f(x)/g(x))=(f'(x)g(x)-g'(x)f(x))/(g(x))^2#

in this problem, #f(x)=ln(x)# and #g(x)=1+ln(x)#

#f'(x)=1/x# and #g'(x)=1/x#

#y'(x)=(1/x(1+lnx)-1/x(lnx))/(1+lnx)^2#

#y'(x)=(1/x+lnx/x-lnx/x)/(1+lnx)^2#

#y'(x)=(1/x)/(1+lnx)^2#

#y'(x)=1/(x(1+lnx)^2)#