The divergence of the vector field

The vector field given in the image is:

$\overrightarrow{u}=e^x(\cos y\hat{i}+\sin y\hat{j})$

We can identify the components of the vector field $\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$ as follows:

$u_x=e^x\cos y$

$u_y=e^x\sin y$

The divergence of the vector field is defined as:

$\text{div}\overrightarrow{u}=\nabla \cdot \overrightarrow{u}=\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}$

Now, compute the partial derivative of the x-component with respect to x:

$\frac{\partial u_x}{\partial x}=\frac{\partial }{\partial x}\left(e^x\cos y\right)=e^x\cos y$

Next, compute the partial derivative of the y-component with respect to y:

$\frac{\partial u_y}{\partial y}=\frac{\partial }{\partial y}\left(e^x\sin y\right)=e^x\cos y$

Add these two partial derivatives to find the total divergence:

$\nabla \cdot \overrightarrow{u}=e^x\cos y+e^x\cos y=2e^x\cos y$

Therefore, the correct option is 2e^x cos y.