Answer
$g'(x) = \frac{1}{1+x^2}$
Work Step by Step
$f(g(x)) = x$
$f'(g(x)) = \frac{d}{dx}(x) = 1$
Also:
$f'(g(x)) = f'(g(x))\cdot g'(x)$
It is given in the question that:
$f'(x) = 1+[f(x)]^2$
Therefore:
$f'(g(x)) = (1+[f(g(x))]^2)\cdot g'(x) = 1$
$g'(x) = \frac{1}{1+[f(g(x))]^2}$
$g'(x) = \frac{1}{1+x^2}$
