$f(x) = 3{x^2} + 1$ f'(x) is bijective function and $f(g(x)) = x \Rightarrow g(x)$ is inverse of f(x) $g(f(x)) = x$ $g'(f(x))\,.\,f'(x) = 1$ $g'(f(x)) = {1 \over {3{x^2} + 1}}$ Put x = 4 we get $g'(63) = {1 \over {49}}$