TITLE: How to show the derivative by using the limit definition?
QUESTION [12 upvotes]: 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!

REPLY [3 votes]: 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.