Derivative of gradient of norm squared
Solution 1:
The key step seems to be an initial application of the Leibniz integral rule. We have $$ \frac{d}{ds} \int_{\Omega}|\nabla (u(x) + sv(x))|^2 dx = \\ \int_{\Omega}\frac{d}{ds} |\nabla u(x) + s \nabla v(x)|^2 dx =\\ \int_{\Omega}\frac{d}{ds} (|\nabla u(x)|^2 + s^2|\nabla v(x)|^2 + 2s\nabla u(x)^T \nabla v(x)) dx =\\ 2 \int_\Omega [s|\nabla v(x)|^2 + \nabla u(x)^T \nabla v(x)]dx. $$ Evaluating this at $s = 0$ gives you the desired result.