Let $\Phi (u)=\tilde{M}\Big( \int_{\Omega}\frac{1}{p(x)}(|\nabla u(x)|^{p(x)}+\alpha (x)|u(x)|^{p(x)})dx\Big)$ where $u\in W^{1,p(x)}(\Omega)$ (the generalized Sobolev space), $\Omega \subset \mathbb{R}^N$, $p\in C(\overline{\Omega})$, $\alpha \in L^{\infty}(\Omega)$ and $\tilde{M}(t)=\int_0^t M(\xi)d\xi$ where $M:(0,\infty) \to \mathbb{R}$ is a continuous function.
How can I show that $\Phi$ is Frechet differentiable whose differential at the point $u\in W^{1,p(x)}(\Omega)$ is $$\Phi' (u)(v)=M\Big( \int_{\Omega}\frac{1}{p(x)}(|\nabla u(x)|^{p(x)}+\alpha (x)|u(x)|^{p(x)})dx\Big)$$ $$\int_{\Omega}(|\nabla u(x)|^{p(x)-2}\nabla u(x)\nabla v(x)+\alpha (x)|u(x)|^{p(x)-2}u(x)v(x))dx,$$ for every $v\in W^{1,p(x)}(\Omega)$.
First define the functional $$ \xi(u) = \int_{\Omega}\frac{1}{p(\mathbf{x})} \left[ \| \nabla u(\mathbf{x}) \|^{p(\mathbf{x})} + \alpha(\mathbf{x}) u(\mathbf{x})^{p(\mathbf{x})} \right] d\mathbf{x} $$
Define now the scalar-valued function $g(a) =\int_0^{\xi(u+av)} M(t) dt$.
The Leibniz integral rules implies that $$ g'(a) =M[\xi(u+av)] \frac{d}{da}\xi(u+av) $$ The second product term writes \begin{eqnarray} (2) &=& \int_{\Omega} d\mathbf{x} \frac{1}{p} \frac{d}{da} \left[ \alpha \cdot (u+av)^{p(\mathbf{x})} + \| \nabla (u+av) \|^{p} \right] \end{eqnarray} where we omit the dependence in $\mathbf{x}$.
The first term in the integrand is now computed. Denote the scalar $c = u+av$, and $e=c^p$. Thus $de = p c^{p-1} dc = p c^{p-1} v da $
The second term in the integrand is now computed. Denote the scalar $c = u+av$, the vector $\mathbf{e} =\nabla [u+av]$, and the scalar $f = \| \mathbf{e} \|^p$ It follows that $df = p \| \mathbf{e} \|^{p-1} d\| \mathbf{e} \|$ Since $d\| \mathbf{e} \|^2 =2 \| \mathbf{e}\| d\| \mathbf{e} \| =2 \mathbf{e} :d\mathbf{e} =2 \mathbf{e} :\nabla [v] da $ we can deduce $df = p \| \mathbf{e} \|^{p-2} \mathbf{e} :\nabla v da$.
These two results give \begin{eqnarray} (2) &=& \int_{\Omega} d\mathbf{x} \left[ \alpha (u+av)^{p-1}v + \| \nabla [u+av] \|^{p-2} \nabla [u+av]:\nabla v \right] \end{eqnarray}
Because \begin{eqnarray} \phi'(u)(v) &=&g'(0)\\ &=&M[\xi(u)] \frac{d}{da}\xi(u+av) \lvert_{a=0}\\ &=&M[\xi(u)] \cdot \int_{\Omega} \left[ \alpha u^{p-1}v + \| \nabla u \|^{p-2} \nabla u:\nabla v \right] d\mathbf{x} \end{eqnarray} which can be arranged in the desired form