\(\eqalign{
& y' = {\rm{[}}\ln (x + \sqrt {1 + {x^2}} ){\rm{]'}} \cr
& {\rm{ = }}{{(x + \sqrt {1 + {x^2}} )'} \over {x + \sqrt {1 + {x^2}} }} = {{1 + {x \over {\sqrt {1 + {x^2}} }}} \over {x + \sqrt {1 + {x^2}} }} = {1 \over {\sqrt {1 + {x^2}} }} \cr} \)
