## Step 1: Express the Unit Vectors in Terms of Cartesian Coordinates In cylindrical coordinates, the unit vectors are given by: \[ \hat{e}_\rho = \cos\phi\,\hat{i} + \sin\phi\,\hat{j} \] \[ \hat{e}_\phi = -\sin\phi\,\hat{i} + \cos\phi\,\hat{j} \] **Explanation:** This expresses the cylindrical unit vectors \(\hat{e}_\rho\) and \(\hat{e}_\phi\) in terms of Cartesian unit vectors \(\hat{i}\) and \(\hat{j}\). This is necessary because derivatives with respect to \(\phi\) are easier to evaluate in this basis. --- ## Step 2: Compute the Derivatives with Respect to \(\phi\) Take the partial derivatives with respect to \(\phi\): \[ \frac{\partial \hat{e}_\rho}{\partial \phi} = -\sin\phi\,\hat{i} + \cos\phi\,\hat{j} = \hat{e}_\phi \] \[ \frac{\partial \hat{e}_\phi}{\partial \phi} = -\cos\phi\,\hat{i} - \sin\phi\,\hat{j} = -\hat{e}_\rho \] **Explanation:** Taking derivatives, we see that the derivative of \(\hat{e}_\rho\) with respect to \(\phi\) gives \(\hat{e}_\phi\), and the derivative of \(\hat{e}_\phi\) with respect to \(\phi\) gives \(-\hat{e}_\rho\). --- ## Final Answer \[ \boxed{ \frac{\partial \hat{e}_\rho}{\partial \phi} = \hat{e}_\phi, \qquad \frac{\partial \hat{e}_\phi}{\partial \phi} = -\hat{e}_\rho } \] All other first derivatives of the cylindrical unit vectors with respect to the coordinates (\(\rho\), \(z\)) are zero.