If $g$ is the inverse of a function $f$ and $f'(x) = \frac{1}{1 + x^5}$ , then $g'(x)$ is equal to ${1 + \{g(x)\}^5}$ . If $g$ is the inverse of a function $f$ and $f'(x) = \frac{1}{1 + x^5}$ , then $g'(x)$ is equal to :
${1 + \{g(x)\}^5}$ .
It is evident that $f'(x)$ will not exist at $x = -1$.
My attempt:  I have used the following Theorem.
TH:  Let $I$ be an interval and $f: I \to \mathbb R$ be the function inverse to $f$. If $f$ s differentiable on $I$ and $f'(x) \neq 0$ for $x \in I$ , the $g$ is differentiable on $J$ and $g' = \frac{1}{f' o g} $
Now $\ f'(x) = \frac{1}{1 + x^5}$.
So we can say $f(x)$ is monotonic increasing on $(-1 , \infty)$ and monotonic decreasing on $(-\infty , -1)$.
Then the inverse function to $f(x)$ , $g(x)$ is differentiable on  $(-1 , \infty) \cup (-\infty , -1)$.
And 
$ g'(x) = \frac{1}{f'(g(x))}$ = $\frac{1}{\frac{1}{1 + {(g(x))}^5}}  = 1 + \{g(x)\}^5$ for all $x \in (-1 , \infty) \cup (-\infty , -1)$. 
Have I gone wrong anywhere? Can anyon eplease check it? And what will happen at the point $x =-1$.
 A: $1/(1+x^5) \sim 1/(5(1+x))$ as $x \to -1$.  This is a non-integrable singularity: the improper integral $\int_{-1}^a dx/x$ diverges.  Therefore not only does $f'(-1)$ not exist, 
neither does $f(-1)$ (more precisely, 
$\lim_{x \to -1} f(-1) = -\infty$).
Actually, if $f$ is supposed to have an inverse function, it can't be defined on all of $(-\infty, -1) \cup (-1, +\infty)$, because it would take the same value more than once.  You can
have $f$ defined on $(-1,+\infty)$ or on $(-\infty, -1)$.
On the other hand, $\int_a^\infty 1/(1+x^5)\; dx$ and 
$\int_{-\infty}^{-a} 1/(1+x^5)\; dx$ converge, so the range of $f$ is bounded above.  Thus $g$ is only defined on $(-\infty, c)$ for some $c$.
EDIT: Maybe some pictures will help.  If you define $f$ on $(-\infty,-1) \cup (-1,\infty)$, its graph will look like something like this:

Such an $f$ will not have an inverse, because it is not one-to-one: e.g. in the graph above, there are two values of $x$ with $f(x) = 0$, one on each side of $-1$.  If you want an inverse, you'll want to restrict $f$ to either $(-\infty, -1)$ or $(1,\infty)$.  The graph of the inverse $g$ of this $f$ restricted to $(1,\infty)$ looks like this:

But since the range of $f$ is only $(-\infty, c)$ for some $c$ (approximately $1.07$ in this case),
$g$ is only defined on $(-\infty, c)$.
A: If $f$ is such that 
$$
f'(x)=\frac{1}{1+x^5}.
$$
Then for $x<-1\Rightarrow f'(x)<0$ and if $x>-1\Rightarrow f'(x)>0$. Hence 
$$
f(x)=\left\{\begin{array}{cc}
\int^{x}_{c_1}\frac{dt}{1+t^5}\textrm{, }c_1<-1\textrm{, }\forall x<-1\\
\int^{x}_{c_2}\frac{dt}{1+t^5}\textrm{, }c_2>-1\textrm{, }\forall x>-1
\end{array}
\right\}
$$
For $x\in(-\infty,c_1)$ we have $x<c_1\Rightarrow f(x)> 0$. For $x\in(c_2,+\infty)$ we have $x> c_2\Rightarrow f(x)> 0$. Hence for every $c_1,c_2$ near $-1$ and $c_1<-1<c_2$, we have when $x\in(-\infty,c_1)\cup(c_2,+\infty)$ the $f(x)>0$. Hence 
$$
f(x)>0\textrm{, }\forall x\in\textbf{R}-\{-1\}.\tag 1
$$
Also $f$ is strictly dicreasing in $A_1=(-\infty,-1)$ and strictly increasing in $A_2=(-1,+\infty)$. Hence $f$ is defined and continuous in all $\textbf{R}$ and $f(-1)=0$. Also $f$ is differentiatable in all $\textbf{R}$ except at $x=-1$. Hence 
$$
f(x)=\left\{\begin{array}{cc}
\int^{x}_{-1}\frac{dt}{1+t^5}\textrm{, }\forall x<-1\\
\int^{x}_{-1}\frac{dt}{1+t^5}\textrm{, }\forall x>-1\\
0\textrm{, if }x=-1
\end{array}\right\}.\tag 2
$$
Also easily $\lim_{x\rightarrow-\infty}f(x)=\int^{-\infty}_{-1}\frac{dt}{1+t^5}=l_1<\infty$ and $\lim_{x\rightarrow+\infty}f(x)=\int^{+\infty}_{-1}\frac{dt}{1+t^5}=l_2<\infty$. Hence 
$$
l_1>f(x)\geq 0\textrm{, when }-\infty<x\leq -1\textrm{ and }0\leq f(x)<l_2\textrm{, when }-1\leq x<+\infty.
$$
Also when $f\in A_1$, then $-\infty<g(x)\leq -1$, with $0\leq x<l_1$. The inverse of $f$ in $A_2$ is defined in $[0,l_2)\rightarrow[-1,+\infty)$. Using $(2)$ we get:
$$
x=\int^{g(x)}_{-1}\frac{dt}{1+t^5}
$$ 
and satisfies
$$
g'(x)=1+g(x)^5.\tag 3
$$
Note that $f^{(-1)}$ is the catoptric of $f$ with respect to line $y=x$. 
NOTE also that the results and the garph are taken by calculating with Mathematica's $NIntegrate$ which is more reliable than symbolic calculus of Mathematica: 
$$
\textrm{f[x_]:=NIntegrate[1/(1+t^5),{t,-1.001,x}]}\textrm{, when }x<-1.001
$$
and
$$
\textrm{f[x_]:=NIntegrate[1/(1+t^5),{t,-0.99,x}]}\textrm{, when }x>-0.99
$$
The results of Robert Israel as also the results of Mathematica's symbolic calculus are wrong. 
The graph of the function $f$ is given bellow.

