eng_Latn: So, we have exactly the same relation that $\frac{\partial F_{2}}{\partial x} = z - e^{x}sin\ y\ $, which is equal to the partial derivative of $\frac{\partial F_{1}}{\partial y}$.
kan_Knda: ಆದ್ದರಿಂದ, ನಾವು $\frac{\ಭಾಗಶಃ F_{2}}{\ಭಾಗಶಃ x} = z - e^{x}sin\ y\ $, ಇದು $\frac{\ಭಾಗಶಃ F_{1}}{\ಭಾಗಶಃ y}$ನ ಭಾಗಶಃ ಉತ್ಪನ್ನಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ.