Is this the Gateaux differential of $F(u, v)$? Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$ and suppose $X(\Omega) =X_1(\Omega)\times X_2(\Omega)$ be a Banach space. Moreover let $p, q\geq 1$ be two real numbers. Suppose the functional $F:X\to\mathbb{R}$ given by
$$ F(u, v) = \frac{1}{p}\int_{\Omega} P(x, u)\vert\nabla u\vert^{p} dx + \frac{1}{q}\int_{\Omega} Q(x, v)\vert\nabla v\vert^{q} dx -\int_{\Omega} T(x, u, v) dx$$
is differentiable. I would like to find its Gateaux differential $(u, v)$ along the direction $(w, z)$. It is true that it is
$$
\begin{split}
\left\langle dF(u, v), (w, z)\right\rangle & = \int_{\Omega} P(x, u)\vert\nabla u\vert^{p-2}\nabla u\cdot\nabla w dx + \frac{1}{p}\int_{\Omega} P_u(x, u) w \vert\nabla u\vert^{p} dx \\
& \quad +\int_{\Omega} Q(x, v)\vert\nabla v\vert^{q-2}\nabla v\cdot\nabla zdx +\frac{1}{q}\int_{\Omega} Q_v(x, v) z\vert\nabla v\vert^{q}dx  \\ 
& \quad\quad-\int_{\Omega} T_u(x, u, v) w dx -\int_{\Omega} T_v(x, u, v) z dx\quad ?
\end{split}
$$
Could anyone help me to understand if my calculation is right? Thank you in advance!
 A: The best way to proceed in this case is to use the classical definition of functional derivative:
$$
\begin{split}
\langle \mathrm{d}F[u,v],(w,z)\rangle & = \frac{\mathrm{d}}{\mathrm{d}t} F(u+t w,v+t z)\big|_{t=0}\\ 
& = \frac{1}{p} \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} P(x, u + t w) \vert\nabla u +t \nabla w\vert^{p} \mathrm{d}x \Big|_{t=0} \\
& \quad + \frac{1}{q}\frac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} Q(x, v + t z)\vert\nabla v + t \nabla z)\vert^{q} \mathrm{d}x \Big|_{t=0}\\
& \quad\quad - \frac{\mathrm{d}}{\mathrm{d} t}\int_{\Omega} T(x, u + t w, v+ t z) \,\mathrm{d}x\Big|_{t=0}\\
\end{split}
$$
Computing separately each term we get
$$
\begin{split}
\frac{1}{p} \frac{\mathrm{d}}{\mathrm{d}t}\int_{\Omega} P(x, u + t w) \vert\nabla u +t \nabla w\vert^{p} \mathrm{d}x \Big|_{t=0} & = \frac{1}{p} \int_{\Omega} P_u(x, u + t w)w \vert\nabla u +t \nabla w\vert^{p} \mathrm{d}x \Big|_{t=0}\\
& \quad + \int_{\Omega} P(x, u + t w) \vert\nabla u +t \nabla w\vert^{p-2} \big(\nabla u\cdot\nabla w+t|\nabla w|^2\big)\, \mathrm{d}x \Big|_{t=0}\\
& = \frac{1}{p} \int_{\Omega} P_u(x, u) \vert\nabla u \vert^{p} w\,\mathrm{d}x + 
\int_{\Omega} P(x, u ) \vert\nabla u \vert^{p-2} \nabla u\cdot\nabla w\, \mathrm{d}x\\
\\
\frac{1}{q}\frac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega} Q(x, v + t z)\vert\nabla v + t \nabla z)\vert^{q} \mathrm{d}x \Big|_{t=0} 
& = \frac{1}{q} \int_{\Omega} Q_v(x, v + t z)z \vert\nabla v +t \nabla z\vert^{q} \mathrm{d}x \Big|_{t=0}\\
& \quad + \int_{\Omega} Q(x, v + t z) \vert\nabla v +t \nabla z\vert^{q-2} \big(\nabla v\cdot\nabla w+t|\nabla w|^2\big)\, \mathrm{d}x \Big|_{t=0}\\
& = \frac{1}{q} \int_{\Omega} Q_v(x, v) \vert\nabla v \vert^{q} w\,\mathrm{d}x + 
\int_{\Omega} Q(x, v ) \vert\nabla v \vert^{q-2} \nabla v\cdot\nabla z\, \mathrm{d}x\\
\\
- \frac{\mathrm{d}}{\mathrm{d} t}\int_{\Omega} T(x, u + t w, v+ t z) \,\mathrm{d}x\Big|_{t=0} & = -\int_{\Omega} \big[T_u(x, u + t w, v+ t z)w +T_v(x, u + t w, v+ t z)z\big]\,\mathrm{d}x\Big|_{t=0}\\
& = - \int_{\Omega} T_u(x, u, v)w\,\mathrm{d}x - \int_{\Omega} T_v(x, u , v)z\,\mathrm{d}x
\end{split}
$$
therefore your calculations are correct.
Note. In order to speed up the calculations, I have have used the evaluation, done earlier in this Q&A, of the functional derivative of the $p$-th power of the modulus of the gradient
$$
\frac{\mathrm{d}}{\mathrm{d} t} |\nabla f+t\nabla\varphi|^p =|\nabla f+t\nabla\varphi|^{p-2}\frac{\mathrm{d}}{\mathrm{d}t}\langle\nabla f+t\nabla\varphi,\nabla f+t\nabla\varphi\rangle.
$$
