Question 835437
Let 
{{{f=ln(sqrt(x^2+1)-x)}}}
{{{u=sqrt(x^2+1)-x}}}
{{{f=ln(u)}}}
Then,
{{{df/dx=(df/du)(du/dx)}}}
{{{df/dx=(d(ln(u))/du)(du/dx)}}}
{{{df/dx=(1/u)(du/dx)}}}
.
.
.
{{{u=(x^2+1)^(1/2)-x}}}
{{{du/dx=(1/2)(x^2+1)^(-1/2)*(2x)-1}}}
{{{du/dx=x/sqrt(x^2+1)-1}}}
.
.
{{{df/dx=(1/(sqrt(x^2+1)-x))*(x/sqrt(x^2+1)-1)}}}