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)=f \nabla g+g \nabla f

Passo 1

O enunciado pede para derivar a regra do gradiente do produto de duas funções escalares f(x,y,z) e g(x,y,z), ou seja, mostrar que
\nabla(fg)=f\,\nabla g + g\,\nabla f.
Nós vamos usar a definição componente a componente do operador gradiente e a regra do produto para derivadas parciais.

Passo 2

Pela definição, o gradiente de fg é o vetor cujas componentes são as derivadas parciais de fg:
\nabla(fg) =\frac{\partial}{\partial x}(f\,g)\,\mathbf{i} +\frac{\partial}{\partial y}(f\,g)\,\mathbf{j} +\frac{\partial}{\partial z}(f\,g)\,\mathbf{k}.

Passo 3

Aplicamos a regra do produto para derivadas parciais a cada componente. Por exemplo,
\frac{\partial}{\partial x}(f\,g) =f\,\frac{\partial g}{\partial x} +g\,\frac{\partial f}{\partial x},
e analogamente
\frac{\partial}{\partial y}(f\,g) =f\,\frac{\partial g}{\partial y} +g\,\frac{\partial f}{\partial y}, \quad \frac{\partial}{\partial z}(f\,g) =f\,\frac{\partial g}{\partial z} +g\,\frac{\partial f}{\partial z}.

Passo 4

Substituindo essas expressões em \nabla(fg) e agrupando termos, obtemos
\nabla(fg) =\bigl(f\,\tfrac{\partial g}{\partial x}+g\,\tfrac{\partial f}{\partial x}\bigr)\mathbf{i} +\bigl(f\,\tfrac{\partial g}{\partial y}+g\,\tfrac{\partial f}{\partial y}\bigr)\mathbf{j} +\bigl(f\,\tfrac{\partial g}{\partial z}+g\,\tfrac{\partial f}{\partial z}\bigr)\mathbf{k}.
Agora distribuindo,
\nabla(fg) =f\bigl(\tfrac{\partial g}{\partial x}\mathbf{i}+\tfrac{\partial g}{\partial y}\mathbf{j}+\tfrac{\partial g}{\partial z}\mathbf{k}\bigr) +g\bigl(\tfrac{\partial f}{\partial x}\mathbf{i}+\tfrac{\partial f}{\partial y}\mathbf{j}+\tfrac{\partial f}{\partial z}\mathbf{k}\bigr),
que é exatamente
\nabla(fg)=f\,\nabla g + g\,\nabla f.

Resposta

\nabla(fg)=f\,\nabla g + g\,\nabla f.

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