Derivative of $t \mapsto \Vert f+tg \Vert_p^p$ 
Suppose $(X,\mathcal A, \mu)$ is a measure space and let $f,g\in L^p(X)$ be real-valued functions, $p\in(1,+\infty)$. Let us define 
  $$
F:\mathbb{R} \ni t \mapsto \int_X \vert f(x)+tg(x) \vert^p d\mu = \Vert f+tg \Vert_p^p
$$
  Prove that $F$ is differentiable and compute $F'(0)$. 

Have you got any ideas? I've tried different things, quite unsuccessfully. I think that the derivability of $F$ is an application of dominated convergence theorem (but I can't see how exactly). 
What about $F'(0)$? 
$$
\lim_{t \to 0}\frac{F(t)-F(0)}{t}=\lim_{t \to 0} \frac{\int_X \vert f+tg\vert^p-\vert f \vert^pd\mu}{t}
$$
but I do not know how to go on. 
Thanks in advance for your kind help.
 A: It is enough to show that the function $F$ is differentiable on every open subset of $\mathbb{R}$. So let $r>0$, and 
$$
\phi: X\times(-r,r) \to [0,\infty],\ \phi(x,t)=s(f(x)+tg(x))=:\phi^x(t),
$$
where 
$$
s: \mathbb{R} \to [0,\infty),\ s(t)=|t|^p.
$$
Since $s$ is differentiable, and
$$
s'(t)=\begin{cases}
p|t|^{p-2}t &\text{ for } t \ne 0\\
0 &\text{ for } t=0
\end{cases},
$$
it follows that for every $x$  in 
$$
\Omega:=\{x \in X:\ |f(x)|<\infty\}\cap\{x \in X:\ |g(x)|<\infty\}
$$ 
the function $\phi^x$ is differentiable and 
$$
(\phi^x)'(t)=\partial_t\phi(x,t)=g(x)s'(f(x)+tg(x)) \quad \forall\ t \in (-r,r).
$$
Therefore 
$$
|\partial_t\phi(x,t)| \le G_r(x):=\max(1,r^{p-1})|g(x)|(|f(x)|+|g(x)|)^{p-1}  \quad \forall\ (x,t) \in \Omega\times(-r,r)
$$
Thanks to Hölder's inequality  we have
$$
\int_XG_r\,d\mu=\max(1,r^{p-1})\int_X|g|(|f|+|g|)^{p-1}\le \max(1,r^{p-1})\|g\|_{L^p(X)}\|(|f|+|g|)\|^{p-1}_{L^p(X)},
$$
i.e. $G_r \in L^1(X)$
Given $t_0 \in (-r,r)$ and a sequence $\{t_n\} \subset (-r,r)$ with $t_n \to t_0$ we set
$$
\tilde{\phi}_n(x,t_0)=\frac{\phi(x,t_0)-\phi(x,t_n)}{t_0-t_n} \quad \forall x \in \Omega, n \in \mathbb{N}. 
$$
Then
$$
\lim_n\tilde{\phi}(x,t_0)=\partial_t\phi(x,t_0) \quad \forall\ x\in \Omega.
$$
Thanks to the MVT there is some $\alpha=\alpha(t_0,t_n) \in [0,1]$ such that
$$
|\tilde{\phi}_n(x,t_0)|=|\partial_t\phi(x,\alpha t_0+(1-\alpha)t_n)|\le G_r(x) \quad \forall\ x \in \Omega, n \in \mathbb{N}.
$$
Applying the dominated convergence theorem to the sequence $\{\tilde{\phi}_n\}$ we get for every $t_0 \in (-r,r)$:
$$
\int_X\partial_t\phi(x,t_0)d\mu(x)=\int_X\lim_n\tilde{\phi}_n(x,t_0)d\mu(x)=\lim_n\int_X\tilde{\phi}_n(x,t_0)=\lim_n\frac{F(t_0)-F(t_n)}{t_0-t_n}=
F'(t_0).
$$
In particular we have
$$
F'(0)=\int_X g(x)s'(f(x))=\int_X fg|f|^{p-2}.
$$
A: Hint: let $F(t,x):=|f(x)+tg(x)|^p$. Check that we can take the derivative under the integral.  
We have $\partial_tF(t,x)=|g(x)|\cdot |f(x)+tg(x)|^{p-1}\operatorname{sgn}(f(x)+tg(x))$, which is, locally in $t$, bounded in $t$ by an itnegrable function of $x$. 
