Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 802: 64

Using the definition of gradient, we verify the following linearity relations: (a) $\nabla \left( {f + g} \right) = \nabla f + \nabla g$ (b) $\nabla \left( {cf} \right) = c\nabla f$

(a) Let $f$ and $g$ be functions of $x$ and $y$. Note that the methods here are applicable to the case where $f$ and $g$ are functions of $x$, $y$ and $z$. We have the gradients of $f$ and $g$: $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right)$, ${\ \ \ }$ $\nabla g = \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right)$ Let $h=f+g$. So, the gradient of $h$ is $\nabla \left( {f + g} \right) = \left( {\frac{{\partial \left( {f + g} \right)}}{{\partial x}},\frac{{\partial \left( {f + g} \right)}}{{\partial y}}} \right)$ $ = \left( {\frac{{\partial f}}{{\partial x}} + \frac{{\partial g}}{{\partial x}},\frac{{\partial f}}{{\partial y}} + \frac{{\partial g}}{{\partial y}}} \right)$ $ = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) + \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right)$ Hence, $\nabla \left( {f + g} \right) = \nabla f + \nabla g$. (b) Let $h = cf$, where $c$ is a constant. So, the gradient of $h$ is $\nabla \left( {cf} \right) = \left( {\frac{{\partial \left( {cf} \right)}}{{\partial x}},\frac{{\partial \left( {cf} \right)}}{{\partial y}}} \right) = c\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right)$ Hence, $\nabla \left( {cf} \right) = c\nabla f$.