Exercise about Newton´s Method I´m study numerical methods and some applications of calculus, and I need some help here: 
Consider $f$ and $g$ are real functions differentiable such that $g'(x)\neq 0$ for all $x$
$a)$ Show that $f$ and $g(f(x))$ has the same critical points, i.e., $f'$ and $g'(f(x))$ has the same roots.
$b)$ If $f(x) = \dfrac{e^x}{x^2+1}$, show that 1 is a critical point of $f$.
$c)$ Apply Newton´s Method for $f'(x)$ with $x_0 = 1.1$ and precision $\mathcal{E_1} = \mathcal{E_2} = 10^{-2}$. (Hint: Use part $a$).
$d)$ Define a sequence of approximation for a critical point of $h(x) = x-\ln(x^2+1)$, Use last results. 
the parts $a)$ and $b)$ I´m know how to do, but $c)$ I don´t know How I can use the hint and $d)$ I have no idea how I can do. How the last results can help for $d)$?
Anyone can help me?
 A: By Newton Method you must find roots of $f'(x)$ function (critical point)
The Newton method say that: $x(n+1) = x(n) - \dfrac{f(x(n))}{f'(x(n))}$. But you must find it for $f'(x)$ function (not for $f(x)$). So you solution is $x(n+1) = x(n) - \dfrac{f'(x)}{f''(x)}$.
Step 1: $x(1) = x(0) - \dfrac{f'(x(0))}{f''(x(0))} = 1.1 - \dfrac{f'(1.1)}{f''(1.1)}$
Step 2: $x(2) = x(1) - \dfrac{f'(x(1))}{f''(x(1))}$
And so on. You will finish make this steps when $\left|{f'(x(n))}\right| <= \epsilon$
In d) I think you must do the same. In tht task you must also make approximation. If I understend right $h(x) = ln(f(x))$ ($f(x)$ as it defined in b))
Because $ln(\dfrac{e^x}{x^2+1}) = ln(e^x) - ln(x^2 + 1) = x - ln(x^2 + 1)$
So, in c) task you will define some $f'(x)$ and in d) - you can use this calculation to define $h'(x) = (ln(f(x)))' = \dfrac{f'(x)}{f(x)}$
A: For 1:
$(g(f(x))'
=f'(x) g'(f(x))
$.
Since
$g'(z) \ne 0$
for any $z$,
$(g(f(x))' = 0
\iff
f'(x) = 0
$,
so
$g(f(x))$
and
$f(x)$
have the same critical points.
