The divergence of the vector field

The correct option is 2e^x cos y.

Step-by-step Explanation:

From the given image, we can identify the vector field
$$\overrightarrow{u}=e^x(\cos y\hat{i}+\sin y\hat{j})$$
We can write this vector field in terms of its component functions along the x and y directions:
$$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$$
where:
$$u_x=e^x\cos y$$
and
$$u_y=e^x\sin y$$

The divergence of a two-dimensional vector field is defined as:
$$\text{div}\left(\overrightarrow{u}\right)=\nabla ·\overrightarrow{u}=\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}$$

Now, we calculate the required partial derivatives:
1. Differentiating $u_x$ partially with respect to $x$ (treating $y$ as a constant):
$$\frac{\partial u_x}{\partial x}=\frac{\partial }{\partial x}\left(e^x\cos y\right)=e^x\cos y$$
2. Differentiating $u_y$ partially with respect to $y$ (treating $x$ as a constant):
$$\frac{\partial u_y}{\partial y}=\frac{\partial }{\partial y}\left(e^x\sin y\right)=e^x\cos y$$

Substituting these partial derivatives back into the divergence formula:
$$\nabla ·\overrightarrow{u}=e^x\cos y+e^x\cos y=2e^x\cos y$$
Thus, the divergence of the vector field is $2e^x\cos y$.