Suppose I have a vector 
such as 
.
Now the divergence of this vector will be
![\[\text{div}~\textbf{A} = \nabla . \textbf{A} = \frac{\partial a_x}{\partial x} + \frac{\partial a_y}{\partial y} + \frac{\partial a_z}{\partial z}.\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image003.webp)
So if I use the technique for first-order partial differentiation of functions with three variables, I will get the divergence of the vector.
If interested, you can read more about the other posts in vector analysis like directional derivative, the gradient of a scalar field, unit normal vector, unit tangent vector, curl of any vector and so on.
Now I’ll give some examples on the divergence of a vector function.
EXAMPLE
According to Kreyszig (2005)*, “Find the divergence of the following vector function: ![[e^{2x}\cos 2y,~~~e^{2x}\sin 2y,~~~5e^{2z}]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image004.webp)
.”
SOLUTION
Now here the given vector is .
First of all, I’ll give it a name, say, .
So, in vector form, it will be
![\[\textbf{A} = e^{2x}\cos 2y\textbf{i}+e^{2x}\sin 2y\textbf{j}+5e^{2z}\textbf{k}.\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image005.webp)
As per the the formula for the divergence of any vector, divergence of will be
![\[\text{div}\textbf{A}= \nabla . \textbf{A} = \frac{\partial}{\partial x}(e^{2x}\cos 2y) + \frac{\partial}{\partial y} (e^{2x}\sin 2y) + \frac{\partial}{\partial z} (5e^{2z}).\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image006.webp)
Thus it will be
![\[\nabla . \textbf{A} = \cos 2y\frac{\partial}{\partial x}(e^{2x}) + e^{2x}\frac{\partial}{\partial y} (\sin 2y) + 5\frac{\partial}{\partial z} (e^{2z}).\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image007.webp)
So this means
![\[\nabla . \textbf{A} = \cos 2y (2.e^{2x}) + e^{2x} (2\cos 2y) + 5 (2. e^{2z}).\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image008.webp)
Now I’ll simplify it to get
![\[\nabla . \textbf{A} = 2e^{2x}\cos 2y + 2 e^{2x} \cos 2y + 10 e^{2z}\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image009.webp)
which means
![\[\nabla . \textbf{A} = 4e^{2x}\cos 2y + 10 e^{2z}.\]](https://appassets.softecksblog.in/engineering_mathematics/assets/em2/16_files/image010.webp)
Hence I can conclude that this is the solution to the given example.
