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(k f)=k \nabla f

Passo 1

O enunciado pede para mostrarmos a propriedade de linearidade do operador gradiente em relação a uma constante. Em outras palavras, queremos provar que, dado um escalar constante k e uma função f(x,y,z), vale
\nabla\bigl(k\,f\bigr)=k\,\nabla f.
Para isso, usaremos a regra escalar
\frac{\partial}{\partial x}\bigl(k\,f\bigr)=k\,\frac{\partial f}{\partial x},
e de modo análogo para as variáveis y e z.

Passo 2

Começamos escrevendo o gradiente de k\,f em componentes:
\nabla\bigl(k\,f\bigr) =\frac{\partial (k\,f)}{\partial x}\,\mathbf{i} +\frac{\partial (k\,f)}{\partial y}\,\mathbf{j} +\frac{\partial (k\,f)}{\partial z}\,\mathbf{k}.
Aplicando a regra de derivada de um produto com constante em cada componente, obtemos
\frac{\partial (k\,f)}{\partial x} = k\,\frac{\partial f}{\partial x},\quad \frac{\partial (k\,f)}{\partial y} = k\,\frac{\partial f}{\partial y},\quad \frac{\partial (k\,f)}{\partial z} = k\,\frac{\partial f}{\partial z}.
Substituindo de volta,
\nabla\bigl(k\,f\bigr) =k\,\frac{\partial f}{\partial x}\,\mathbf{i} +k\,\frac{\partial f}{\partial y}\,\mathbf{j} +k\,\frac{\partial f}{\partial z}\,\mathbf{k} =k\,\bigl( \frac{\partial f}{\partial x}\,\mathbf{i} +\frac{\partial f}{\partial y}\,\mathbf{j} +\frac{\partial f}{\partial z}\,\mathbf{k} \bigr).

Resposta

\nabla\bigl(k\,f\bigr)=k\,\nabla f.

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