How to show the derivative by using the limit definition? Let $(\Omega, \mathcal{F}, \mu)$ be a probability space. For the sake of simplicity, let $\Omega$ be $\mathbb{R}$. In what follows, the measurability always refers to Borel measurability.
Let $f \colon \mathbb{R}_+ \times \Omega \to \mathbb{R}_+$ be a function such that:
(i) For each $z \in \Omega$, the function $k \mapsto f(k,z)$ is concave, increasing, and continuously differentiable,
while $z \mapsto f(k,z)$ is Borel measurable for each $k \in \mathbb{R}_+$;
(ii) $\lim_{k \downarrow 0} f'(k,z) >0$ for each $z \in \Omega$.
Here and below, $f'(k,z)$ denotes the partial derivative of $f$ with respect to $k$; and
(iii) $f(0,z)=0$ for all $z \in \Omega$.
Let $v \colon \mathbb{R}_+ \to \mathbb{R}_+$ be a bounded, strictly concave and strictly increasing function,
and be continuously differentiable on $(0, \infty)$.
Define a function $g $  by
\begin{align*}
g(k) := \left( \int_{\Omega} \left[ v\left( f( k, z ) \right) \right]^\alpha \mu (\mathrm{d}z) \right)^{1/\alpha},
\qquad (0<\alpha <1).
\end{align*}

Question:
In fact, since $g$ is concave (it has been proved), we know that the right-hand and the left-hand derivatives of $g$ exist. 
I aim to show that $g$ is differentiable on $(0, \eta)$ for any fixed constant $\eta >0$, and to show the derivative of $g$ which I conjecture is
  \begin{align*}
g'(k) =  \left( \int_{\Omega} \left[ v\left( f( k, z ) \right) \right]^\alpha \mu (\mathrm{d}z) \right)^{\frac{1}{\alpha} -1 }  
\int_{\Omega} \left[ v\left( f( k, z ) \right) \right]^{\alpha -1} 
v'\left( f( k, z ) \right)
f'( k, z ) \mu( \mathrm{d} z)
\end{align*}
  for all $0 < k < \eta$.

Here, $g'(k) = \dfrac{\mathrm{d}}{\mathrm{d} k} g(k)$, $v'\left( f( k, z ) \right) := \dfrac{\mathrm{d}}{\mathrm{d} f }v( f( k, z ))$,
and $f'(k, z) := \dfrac{\partial}{\partial k} f(k,z)$.
My attempt:

The above stated derivative of $g$ is just my conjecture and I am not sure if the right-hand derivative of $g$ is equal to its left-hand derivative, thus I wish to verify it.
  My attempt is making use of the limit definition to show the left-hand and right-hand derivatives of $g$ are the same and equal to the above stated formula.

In fact, I even got stuck in finding the left-hand side and the right-hand side derivatives of $g$. 
But I thought it might suffice to show that the left-hand side derivative $g’_-(k) := \lim_{h \to 0^-} \dfrac{g(k+h)-g(k)}{h}$ is less than the conjecture formulation that stated above, and to show that the right-hand side derivative $g’_+(k) := \lim_{h \to 0^+}\dfrac{g(k+h)-g(k)}{h}$  is greater than the conjecture formulation. Then, by concavity of $g$, we have $g’_-(k) \geq g’_+(k)$ and hence, $g’_-(k)= g’_+(k)=g’(k)$ as desired. In this connection, I think the problem becomes how to establish the relation between the right-hand derivative and conjecture formula, and relation between the left-hand derivative and conjecture formula. 
Could anyone give me some guidance and help me out please?
Thank you very much in advance!
 A: To use the Liebnitz rule, nicely explained by Vlad, we need
$$\tag 1 \int_\Omega \partial_k (v^\alpha \circ f(k,z))\,d \mu (z) <\infty.$$
Think about this for a bit: If we had $\Omega, f, v,$ such that the integrand is not bounded for some $k,$ we could fashion a probability measure $\mu$ on $\Omega$ for which $(1)$ fails. So boundedness is something we need to look for.
Let $\mathcal U$ be the collection of functions $u\in C^1(0,\infty)$ that are positive, increasing, and concave. Let $\mathcal U_b$ be the set of bounded functions $u\in \mathcal U.$
Lemma: Suppose $u\in \mathcal U.$  Then
$$u'(x) \le \frac{u(x)}{x}\,\,\text {for all } x>0.$$
If in addition $u\in \mathcal U_b,$ then 
$$u'(x)x \le \|u\|_\infty.$$
I'll leave the proof of this to you for now. It's fairly simple, but ask questions if you like.
Let's turn to $(1).$ We are given $v\in \mathcal U_b.$ This implies $v^\alpha \in \mathcal U_b.$ Thus by the lemma, we have
$$ (v^\alpha)'(x) x \le  \|v^\alpha\|_\infty.$$
We also have $f(k,z) \in \mathcal U$ as a function of $k$ for each fixed $z.$ Thus the lemma gives
$$f'(k,z) \le \frac{f(k,z)}{k}.$$
It follows that
$$\tag 2(v^\alpha)'(f(k,z)) f'(k,z) \le (v^\alpha)'(f(k,z)) \frac{f(k,z)}{k} \le \frac{\|v^\alpha\|_\infty}{k}.$$
But the left side of $(2)$ is exactly the integrand in $(1)$. Thus for $k \ge k_0$ we have the uniform bound $\|v^\alpha\|_\infty/k_0$ on the integrands. Of course the constant function equal to to this bound belongs to $L^1(\mu),$ since we're in a probablity space. This allows us to use Leibnitz on each interval $(k_0,\infty),$ and gives the desired result.
A: Perhaps I am missing something out, but it seems to me that you can obtain your final result using Leibniz's rule as shown below.
$$
\begin{align*}
g(k) := \left( \int_{\Omega} \big[ v\left( f( k, z ) \right) \big]^\alpha \mu (\mathrm{d}z) \right)^{\frac{1}{\alpha}}
=I^\frac{1}{\alpha}\left(k\right)\qquad (0<\alpha <1).
\end{align*}
$$
where 
$$
I(k)=\int_{\Omega} v^\alpha \left( f( k, z ) \right) \mu (\mathrm{d}z)
$$
Thus, if we apply regular chain rule we get
$$
\frac{\partial }{\partial k}\left[g\left(k\right)\right] =\frac{\partial }{\partial k}\left[I^\frac{1}{\alpha}\left(k\right)\right]
=\frac{1}{\alpha}I^{\frac{1}{\alpha}-1}\left(k\right) \frac{\partial }{\partial k}\left[I \left(k\right)\right]
$$
Then,  using Measure theory form of Leibniz rule we write
$$
 \frac{\partial }{\partial k}\left[I \left(k\right)\right] =
 \frac{\partial }{\partial k}\left[\int_{\Omega} \big[ v\left( f( k, z ) \right) \big]^\alpha \mu \left(\mathrm{d}z\right)\right] =
\int_{\Omega} \frac{\partial }{\partial k}\big[ v\left( f( k, z ) \right) \big]^\alpha \mu\left(\mathrm{d}z\right) =
\int_{\Omega} \alpha v^{\alpha-1}\left( f( k, z ) \right) v'\left( f( k, z )\right) \frac{\partial }{\partial k}\big[  f( k, z )  \big] \mu (\mathrm{d}z) =
\alpha\int_{\Omega}  v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right)   \mu \left(\mathrm{d}z\right)
$$
Substituting the last expression into previous formula for $g'(k)$ we get
$$
g'\left(k\right) = \frac{1}{\alpha} \left(\int_\Omega \nu^\alpha\left(f\right)\mu\left(\mathrm{d}z\right)\right)^{\frac{1}{\alpha}-1} 
\cdot
\alpha\int_{\Omega}  v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right)   \mu \left(\mathrm{d}z\right)
$$
$$
\boxed{g'\left(k\right) =  \left(\int_\Omega \nu^\alpha\left(f\right)\mu\left(\mathrm{d}z\right)\right)^{\frac{1}{\alpha}-1} 
\int_{\Omega}  v^{\alpha-1}\left( f \right) v'\left( f \right) f_k\left( k, z \right)   \mu \left(\mathrm{d}z\right)}
$$
where $f=f\left(k,z\right)$ and $f_k\left(k,z\right)=\dfrac{\partial f\left(k,z\right)}{\partial k}$.

Justification for Using Leibniz's rule
As pointed out in comments by the OP, I have not provided justification for using Leibniz's rule in measure-theoretical form:

Let  $X$ is an open subset of $  \mathbb{R} $, and $\Omega$  is a measure space. 
  Suppose $ F\colon X\times \Omega \rightarrow \mathbb {R}$ satisfies the following conditions:
  
  
*
  
*$ F(x,\omega )$ is a Lebesgue-integrable function of $\omega$  for each $ x\in X$.
  
*For almost all $\omega \in \Omega$, the derivative $ F_{x} $ exists for all $ x\in X$.
  
*There is an integrable function $\displaystyle \theta \colon \Omega \rightarrow \mathbf {R}$ such that $$\left\lvert F_{x}(x,\omega )\right\rvert\leq \theta (\omega )$$  for all $ x\in X$ and almost every $\omega \in \Omega $.
Then, for all $x\in X$,
  $$
{\frac {d}{dx}}\int _{\Omega }F(x,\omega )\,d\omega =\int _{\Omega }F_{x}(x,\omega )\,d\omega .
$$

Let us verify the last condition for the integrand
$$
F\left(k,z\right)=\left[ v\big( f( k, z ) \big) \right]^\alpha , \qquad 0<\alpha<1.
$$
Observe that $F\left(k,z\right)$ is superposition of three concave increasing functions (treating $k\mapsto f\left(k,z\right)$ as function of single variable $k$)
$$
F\left(\cdot\right) = F_1\circ F_2\circ F_3 = F_3\left(F_2\left(F_1\left(\cdot\right)\right)\right),
$$
where 
$$
\begin{aligned}
F_1\left(\tau\right) &= f\left(\tau,z\right), &&\text{where $z$ is fixed} \\
F_2\left(\tau\right) &= v\left(\tau\right), &&\\
F_3\left(\tau\right) &= \tau^{\alpha}, && 0<\alpha<1
\end{aligned}
$$
with $\tau$ being dummy variable.
As an exercise, I propose you to try to show explicitly that superposition of $F\left(\cdot\right) = F_1\circ F_2\circ F_3 =  v^\alpha\left( f\left( k, z \right) \right)$ is convex in $k$.
Recall that concave function of single variable has monotonically decreasing derivative, which means that $F'\left(\tau\right)\leq F'\left(\tau_0\right)$ whenever $\tau\leq\tau_0$.
For example, we can get bound 
$$
\begin{aligned}
F'\left(\tau\right)&<F'\left(0\right)& \text{since}& &f \colon \mathbb{R}_+ \times \Omega \to \mathbb{R}_+ \implies \tau>0
\end{aligned}
$$
where $F'\left(0\right)$ does not depend on $\tau$ (and possibly depends on $\omega$).
Thus, choosing nonnegative constant $k_0$ we can bound derivative as
$$
\frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big] \leq \left.\frac{\partial}{\partial k} \Big[v^\alpha\big( f\left( k, z \right) \big)\Big]\right\rvert_{k_0} 
\qquad\text{ whenever }   k\geq k_0. 
$$
Thus the third condition is satisfied.
