Proof of Identity for Curl of Product of Function With a Vector

\[\mathbf{\nabla} \times (f \mathbf{v})=f \nabla \times \mathbf{v} + (\mathbf{\nabla f}) \times \mathbf{v}\]

\[\begin{equation} \begin{aligned} \mathbf{\nabla} \times (f \mathbf{v}) &= (\frac{\partial (fv_3)}{\partial y}-\frac{\partial (fv_2)}{\partial z}) \mathbf{i} + (\frac{\partial (fv_1) }{\partial z}-\frac{\partial (fv_3 )}{\partial x}) \mathbf{j} + (\frac{\partial (fv_2)}{\partial x}-\frac{\partial (fv_1) }{\partial y}) \mathbf{k} \\ &= (f \frac{\partial v_3}{\partial y} + \frac{\partial f}{\partial y} v_3 -f \frac{\partial v_2}{\partial z}- \frac{\partial f}{\partial z} v_2) \mathbf{i} + (f \frac{\partial v_1}{\partial z}+ \frac{\partial f}{\partial z} v_1 -f \frac{\partial v_3}{\partial x}-\frac{\partial f}{\partial x}v_3 ) \mathbf{j} + (f \frac{\partial v_2}{\partial x} + \frac{\partial f}{\partial x} v_2 - f \frac{\partial v_1}{\partial y}- \frac{\partial f}{\partial y}v_1 ) \mathbf{k} \\ &= f( \frac{\partial v_3}{\partial y} - \frac{\partial v_2}{\partial z}) \mathbf{i} + ( \frac{\partial v_1}{\partial z}- \frac{\partial v_3}{\partial x} ) \mathbf{j} + ( \frac{\partial v_2}{\partial x} - \frac{\partial v_1}{\partial y}) \mathbf{k} \\ &+ (v_3 \frac{\partial f}{\partial y} - v_2 \frac{\partial f}{\partial z}) \mathbf{i} + (v_1 \frac{\partial f}{\partial z} - v_3 \frac{\partial f}{\partial x}) \mathbf{j} + + (v_2 \frac{\partial f}{\partial x} - v_1 \frac{\partial f}{\partial y}) \mathbf{k} \\ &= f \nabla \times \mathbf{v} + (\mathbf{\nabla }f) \times \mathbf{v} \end{aligned} \end{equation}\]

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