$\begin{array}{l}F(x,y) = e^{\displaystyle{2x^2}} - 4y\cos(3y - 2x)\\ +) \quad F'_x = [4x.e^{\displaystyle{2x^2}} +4y\sin(2x - 3y).2]dx\\ \to F'_x = [4x.e^{\displaystyle{2x^2}} +8\sin(2x - 3y)]dx\\ +) \quad F'_y = \{-4[(\cos(2x-3y) -\sin(2x - 3y).(-3)]\}dy\\ \to F'_y = [-4\cos(2x - 3y) - 12\sin(2x - 3y)]dy\\ Vậy\,\,F'(x,y) = [4x.e^{\displaystyle{2x^2}} +8\sin(2x - 3y)]dx + [-4\cos(2x - 3y) - 12\sin(2x - 3y)]dy\end{array}$
