The algebra rules for gradients Given a constant k and the gradients
\nabla f=\frac{\partial f}{\partial x} \mathbf{i}+\frac{\partial f}{\partial y} \mathbf{j}+\frac{\partial f}{\partial z} \mathbf{k}
and
\nabla g=\frac{\partial g}{\partial x} \mathbf{i}+\frac{\partial g}{\partial y} \mathbf{j}+\frac{\partial g}{\partial z} \mathbf{k}
use the scalar equations
\begin{array}{cc} \frac{\partial}{\partial x}(k f)=k \frac{\partial f}{\partial x}, & \frac{\partial}{\partial x}(f \pm g)=\frac{\partial f}{\partial x} \pm \frac{\partial g}{\partial x} \\\ \frac{\partial}{\partial x}(f g)=f \frac{\partial g}{\partial x}+g \frac{\partial f}{\partial x}, & \frac{\partial}{\partial x}\left(\frac{f}{g}\right)=\frac{g \frac{\partial f}{\partial x}-f \frac{\partial g}{\partial x}}{g^{2}} \end{array}
and so on, to establish the following rules.
\nabla(f-g)=\nabla f-\nabla g

Passo 1

O enunciado pede para mostrar que o operador gradiente é linear em relação à subtração de funções, ou seja, provar que
\nabla(f - g) = \nabla f - \nabla g. Vamos usar a definição componente a componente
\nabla h = \frac{\partial h}{\partial x}\mathbf{i} + \frac{\partial h}{\partial y}\mathbf{j} + \frac{\partial h}{\partial z}\mathbf{k}
e a propriedade
\frac{\partial}{\partial x}(f - g) =\frac{\partial f}{\partial x}-\frac{\partial g}{\partial x},
além de suas analogias para y e z.

Passo 2

Aplicando a regra da derivada de diferença à componente x, obtemos
\frac{\partial}{\partial x}(f - g) =\frac{\partial f}{\partial x}-\frac{\partial g}{\partial x}.

Passo 3

De modo análogo para as componentes y e z, temos
\frac{\partial}{\partial y}(f - g) =\frac{\partial f}{\partial y}-\frac{\partial g}{\partial y}, \qquad \frac{\partial}{\partial z}(f - g) =\frac{\partial f}{\partial z}-\frac{\partial g}{\partial z}.

Passo 4

Agora montamos o vetor gradiente de f-g e substituímos as componentes que calculamos:
\nabla(f - g) =\frac{\partial(f-g)}{\partial x}\mathbf{i} +\frac{\partial(f-g)}{\partial y}\mathbf{j} +\frac{\partial(f-g)}{\partial z}\mathbf{k}
\;=\Bigl(\frac{\partial f}{\partial x}-\frac{\partial g}{\partial x}\Bigr)\mathbf{i} +\Bigl(\frac{\partial f}{\partial y}-\frac{\partial g}{\partial y}\Bigr)\mathbf{j} +\Bigl(\frac{\partial f}{\partial z}-\frac{\partial g}{\partial z}\Bigr)\mathbf{k}.
Isso coincide exatamente com
\nabla f - \nabla g.

Resposta

\displaystyle \nabla(f - g) = \nabla f - \nabla g

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