Question 1183465
To show:  {{{(1+x^2)(dy/dx)^2-y^2=0}}}, or {{{(1+x^2)(dy/dx)^2=y^2}}}

{{{y = x-sqrt(1+x^2)}}} ==> {{{dy/dx = 1 - x/sqrt(1+x^2)}}}


==>  {{{(dy/dx)^2 = (1 - x/sqrt(1+x^2))^2 = 1 - (2x)/sqrt(1+x^2) + x^2/(1+x^2)}}}

==> {{{highlight((1+x^2)(dy/dx)^2) = (1+x^2)(1 - x/sqrt(1+x^2))^2 = (1+x^2)( 1 - (2x)/sqrt(1+x^2) + x^2/(1+x^2))

= 1+x^2 -2x*sqrt(1+x^2) + x^2 = x^2 -2x*sqrt(1+x^2) +(1 + x^2) = (x-sqrt(1+x^2))^2 = highlight(y^2)
}}}