>>11672250We start with f'(x) = f^-1(x)
Notice that f'(x) = 1/f'^-1(y) where y = f(x)
f^-1(y) = x(y) but we choose x(y) to be x(x) -> x
therefore 1/f'^-1(y(x)) = x
1 = x*f'^-1(y(x))
rewrite f'^-1(y(x)) as dx/dy and we get:
1 = x*dx/dy -> dy = x*dx
which yields y = f(x) = 1/2x^2
