$f(n) = \int_{1}^{n} n^{x^{-1}}dx$. $\frac{df}{dn}$? Lets start with the domain $1 \leq n \in \mathbb{R}$:
$f(n) = \int_{1}^{n} n^{x^{-1}}dx$
$\frac{df}{dn}$ should be sort of logarithmic and probably not interesting.?
What is it, though?
What have I tried?
To see how far it is from something interesting, and maybe more..
 A: 
METHODOLOGY $1$:  EVALUATION OF $\displaystyle f(n)$ AND DIRECT DIFFERENTIAION

EVALUTING $\displaystyle f(n)$:
We can write the integral of interest in terms of the Exponential Integral $\text{Ei}(x)$.
Note that we have
$$\begin{align}
f(n)&=\int_1^n n^{1/x}\,dx\tag1\\\\
&=\int_{1/n}^1 \frac{n^x}{x^2}\,dx\\\\
&=\int_{1/n}^1 \frac{e^{\log(n)x}}{x^2}\,dx\\\\
&=\log(n)\int_{\log(n)/n}^{\log(n)} \frac{e^x}{x^2}\,dx\\\\
&=\log(n)\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)+n(n^{1/n}-1)\tag 2
\end{align}$$

DIFFERENTIATING $\displaystyle f(n)$:
To determine $f'(n)$ we begin with $(2)$ and write
$$\begin{align}
f'(n)&=\frac1n\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)\\\\
&+\log(n)\left(\frac{1}{\log(n)}-\frac{n}{\log(n)}\frac{dn^{1/n}}{dn}\right)\\\\
&+(n^{1/n}-1)+n\frac{dn^{1/n}}{dx}\\\\
&=n^{1/n}+\frac1n\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)\tag 3
\end{align}$$  


METHODOLOGY $2$:  APPLYING LEIBNIZ'S RULE

Using Leibniz's Rule, we differentiate $(1)$ to obtain
$$\begin{align}
f'(n)&=n^{1/n}+\int_1^n \frac{n^{1/x-1}}{x}\,dx\\\\
&=n^{1/n}+\frac1n \int_{1/n}^1 \frac{n^x}{x}\,dx\\\\
&=n^{1/n}+\frac1n \int_{\log(n)/n}^{\log(n)}\frac{e^x}{x}\,dx\\\\
&=n^{1/n}+\frac1n\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)
\end{align}$$
which agrees with the result reported in $(3)$!
