Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 802: 66

Proof of the Product Rule for Gradients: $\nabla \left( {fg} \right) = f\nabla g + g\nabla f$

Let $f\left( {x,y,z} \right)$ and $g\left( {x,y,z} \right)$ be differentiable and $c$ a constant. If $F\left( t \right)$ is a differentiable function of one variable, then according to Theorem 1, the Product Rule for Gradients is given by $\nabla \left( {fg} \right) = f\nabla g + g\nabla f$ Proof. Using the definition of gradient we get $\nabla \left( {fg} \right) = \left( {\frac{{\partial \left( {fg} \right)}}{{\partial x}},\frac{{\partial \left( {fg} \right)}}{{\partial y}},\frac{{\partial \left( {fg} \right)}}{{\partial z}}} \right)$ $ = \left( {f\frac{{\partial g}}{{\partial x}} + g\frac{{\partial f}}{{\partial x}},f\frac{{\partial g}}{{\partial y}} + g\frac{{\partial f}}{{\partial y}},f\frac{{\partial g}}{{\partial z}} + g\frac{{\partial f}}{{\partial z}}} \right)$ $ = \left( {f\frac{{\partial g}}{{\partial x}},f\frac{{\partial g}}{{\partial y}},f\frac{{\partial g}}{{\partial z}}} \right) + \left( {g\frac{{\partial f}}{{\partial x}},g\frac{{\partial f}}{{\partial y}},g\frac{{\partial f}}{{\partial z}}} \right)$ $ = f\left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}},\frac{{\partial g}}{{\partial z}}} \right) + g\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$ Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$ and $\nabla g = \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}},\frac{{\partial g}}{{\partial z}}} \right)$, hence $\nabla \left( {fg} \right) = f\nabla g + g\nabla f$