The Gradient in Cylindrical Polar Coordinates
\[( \mathbf{i} , \mathbf{j} , \mathbf{k} )\]
and \[( \mathbf{e_r} , \mathbf{e_{\theta}} , \mathbf{e_z} )\]
we can express the vector \[\mathbf{a}\]
as\[\mathbf{a} =a_1 \mathbf{i} +a_2 \mathbf{j} +a_3 \mathbf{k} = b_1 \mathbf{e_r} , b_2 \mathbf{e_{\theta}} + b_3 \mathbf{e_z} \]
We can take the dot product of
\[\mathbf{a}\]
with each of the cylindrical basis vectors gives\[\mathbf{a} \cdot \mathbf{e_r} = b_1\]
\[\mathbf{a} \cdot \mathbf{e_{\theta}} = b_2\]
\[\mathbf{a} \cdot \mathbf{e_z} = b_3\]
Hence
\[\mathbf{a} = (\mathbf{a} \cdot \mathbf{e_r}) \mathbf{e_r} + (\mathbf{a} \cdot \mathbf{e_{\theta}}) \mathbf{e_{\theta}} + (\mathbf{a} \cdot \mathbf{e_z}) \mathbf{e_z}\]
Let
\[\mathbf{a} = \mathbf{\nabla} f\]
then\[\mathbf{{\nabla}} = (\mathbf{{\nabla}} \cdot \mathbf{e_r}) \mathbf{e_r} + (\mathbf{{\nabla}} \cdot \mathbf{e_{\theta}}) \mathbf{e_{\theta}} + (\mathbf{{\nabla}} \cdot \mathbf{e_z}) \mathbf{e_z}\]
Now
\[(ds)^2 = \sqrt{(dr)^2 + r^2 ( d \theta )^2 + (dz)^2} \rightarrow \frac{\partial r}{\partial s}= \frac{\partial z}{\partial s}=1, \: \frac{\partial s}{\partial \theta } = r \rightarrow \frac{\partial \theta}{\partial s}= \frac{1}{r} \]
\[\mathbf{e_r} \cdot (\mathbf{\nabla} f) = \frac{\partial f}{\partial s} |_{\theta , z = constant} = \frac{\partial r}{\partial s} \frac{\partial f}{\partial r} = \frac{\partial f}{\partial r}\]
\[\mathbf{e_\theta} \cdot (\mathbf{\nabla} f) = \frac{\partial f}{\partial s} |_{r , z = constant} = \frac{\partial \theta}{\partial s} \frac{\partial f}{\partial \theta} =\frac{1}{r} \frac{\partial f}{\partial \theta}\]
\[\mathbf{e_z} \cdot (\mathbf{\nabla} f) = \frac{\partial f}{\partial z} |_{r, \theta = constant} = \frac{\partial z}{\partial s} \frac{\partial f}{\partial z} = \frac{\partial f}{\partial z}\]
Hence
\[\mathbf{\nabla} f = \frac{\partial f}{\partial r} \mathbf{e_r} _+ \frac{1}{r} \frac{\partial f}{\partial \theta} \mathbf{e_\theta}+ \frac{\partial f}{\partial z} \mathbf{e_z} \]
