Tricky!

$f\left( x \right) =\int _{ 2 }^{ x }{ \frac { dx }{ \sqrt { 1+{ x }^{ 4 } } } } \\ as\quad g\quad is\quad the\quad inverse\quad of\quad f\quad therefore,\\ g(f\left( x \right) )=x\\ differentiating\quad wrt\quad x\quad \\ g'(f\left( x \right) )\times f'\left( x \right) =1\quad ......(1)\\ f'\left( x \right) =\frac { 1 }{ \sqrt { 1+{ x }^{ 4 } } } \quad ......(2)\\ using\quad (1)\quad and\quad (2)\quad \\ g'(f\left( x \right) )=\sqrt { 1+{ x }^{ 4 } } \\ \\ Now,as\quad it\quad can\quad be\quad easily\quad seen\quad that\quad f\left( x \right) =0\quad only\quad when\quad x=2\\ therefore,\\ g'(f\left( 2 \right) )=\sqrt { 17\quad } \\ g'\left( 0 \right) =\sqrt { 17 } =n\\ { n }^{ 2 }=17$