Is $\frac {e^{\frac 1x} + 1} {e^{\frac 1x}}$ the derivative of some function I need to determine if there exists a differentiable function $f$ in $\mathbb R$ such that for every $x \neq 0$ , $f'(x)=\frac {e^{\frac 1x} + 1} {e^{\frac 1x}}$
I've been pondering this for quite a while, but I really have absolutely no idea how to approach this question. I would really like to hear how one tackles this, what the thought process is, and about different methods/approaches/theorems.
EDIT:
For clarification, this question is part of Calculus I. Although I know about integrals to some extent (from high-school), they haven't been discussed in the course. So this needs to be solved without them. And, according to some comments, I believe what I am after is a closed form "primitive", elementary function.
 A: Note
$$
g(x) = \frac{e^{1/x}+1}{e^{1/x}} = 1+e^{-1/x}
$$
From this we see
$$
\lim_{x \to 0^+} g(x) = 1,\qquad \lim_{x \to 0^-}g(x) = +\infty ,
$$
so $g$ is discontinuous at $0$.  But aside from $0$, the function is continuous.  So we can apply the fundamental theorem of calculus.
$$
f_1(x) = \int_1^x g(x)\;dx
$$
exists for all $x>0$ and $f_1'(x) = g(x)$ for all $x>0$.
And
$$
f_2(x) = \int_{-1}^x g(x)\;dx
$$
exists for all $x<0$ and $f_2'(x) = g(x)$ for all $x<0$.

A "closed form" integral can be written in terms of the exponential integral function.
$$
f(x) = x(1+e^{-1/x}) - \operatorname{Ei}_1(1/x)+C, \quad x>0,
\\
f(x) = x(1+e^{-1/x}) + \operatorname{Ei}(-1/x)+C, \quad x<0.
$$
These are not elementary functions.

There is no function differentiable on all of $\mathbb R$ so that $f'(x) = g(x)$ for all $x \ne 0$.  If $f$ is differentable on on $\mathbb R$ then $f'$ has the intermediate value property.  But $g$ does not have the IVP, even if extended by to have some value $g(0)$.  Indeed, for $x_1>0$ near $0$, we have $g(x_1)<2$, and for $x_2<0$ near $0$ we have $g(x_2)>3$, so some numbers between $f'(x_1)$ and $f'(x_2)$ are not values of $f'$.
A: No, it can't. Check Darboux Theorem: the derivative of a function always has the intermediate value property (IVP).
Since $\;f'(x)\xrightarrow[x\to 0^-]{}\infty\;$ and $\;f'(x)<2\;\;\forall x>0\;$ , and $\;f'(x)\neq 2\;$ for any real $\;x\;$ , this function doesn't fulfill the IVP
A: People are saying yes, people are saying no. I got it wrong at first. One needs to read the question carefully. Say $$V=\Bbb R\setminus\{0\}.$$The answers saying yes show that there is a function defined and differentiable on $V$ with derivative given by that formula. But the problem asks about a function differentiable on $\Bbb R$.
People are saying, correctly it seems to me, that the Darboux theorem shows that there is no such function. Others cite versions of the theorem that are not sufficient (unless they are; the definition of "essential singularity" in a real-variable context is not so clear to me). Yet others don't see why the hypotheses hold...
Let's not cite anything, let's just give a very simple self-contained argument:


Theorem. Suppose $f:\Bbb R\to\Bbb R$ is differentiable. There exist $x_n<0$ with $x_n\to0$ and $f'(x_n)\to f'(0)$.


Proof: Use MVT to find $x_n\in(-1/n,0)$ with $$\frac{f(-1/n)-f(0)}{1/n}=f'(x_n).$$The definition of $f'(0)$ shows that $\frac{f(-1/n)-f(0)}{1/n}\to f'(0)$.
A: Yes, it's the derivative of $f(x)=\int \frac {e^{\frac 1x} + 1} {e^{\frac 1x}} ~dx$ (on a domain that does not include 0).
But I assume you're actually asking if the integral can be expressed in a “closed form” expression with the operators and functions that you learn about in high school.  And the answer to that (at least according to WolframAlpha) is no.  You can only write it in terms of the exponential integral function, or approximate it numerically.
