Let $f \in C^2[a,b]$ so that $f'(x) >0$ for every $ x\in [a,b]$. Solve the equation $f(x) = 0$ by applying Newton's method to the function : $$g(x) =\frac{f(x)}{\sqrt{f'(x)}}$$ Write the Newtion iteration of g and determine the iteration sequence for $f(x) = x^2-2$ and it's order of convergence.
I started solving this by writing: $$x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} = x_n - \frac{f(x_n)}{\sqrt{f'(x_n)}} \frac{2f'(x_n)^{3/2}}{2f'(x_n)^2 - f(x_n)f''(x_n)} = x_n - \frac{2f(x_n)f'(x_n)}{2f'(x_n)^2 - f(x_n)f''(x_n)} $$
If that's ok I have $f(x) = x^2-2%$ and : $f'(x) = 2x$, $f''(x) = 2$, $f'(x)^2 = 4x^2$.
If I plug the values into the relation I wrote above and try to iterate to find the solution $\sqrt{2}$ I get wrong values and that mean I am doing something wrong here. Also how to approach finding the order of converfence for this sequence?