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\left(\frac{f}{g}\right)=\frac{g \nabla f-f \nabla g}{g^{2}}

Passo 1

O enunciado pede para mostrar que, dados dois escalares diferenciáveis f(x,y,z) e g(x,y,z), vale a fórmula
\nabla\Bigl(\frac{f}{g}\Bigr) =\frac{g\,\nabla f - f\,\nabla g}{g^{2}}.
Aqui usamos que
\nabla h = \frac{\partial h}{\partial x}\mathbf{i}+\frac{\partial h}{\partial y}\mathbf{j}+\frac{\partial h}{\partial z}\mathbf{k}
e as regras usuais de derivada parcial (constante, soma, produto e quociente).

Passo 2

Calculamos primeiro a derivada parcial em relação a x. Pela regra do quociente
\frac{\partial}{\partial x}\Bigl(\frac{f}{g}\Bigr) =\frac{g\,\frac{\partial f}{\partial x}-f\,\frac{\partial g}{\partial x}}{g^{2}}.

Passo 3

De modo análogo, para as outras coordenadas obtemos
\frac{\partial}{\partial y}\Bigl(\frac{f}{g}\Bigr) =\frac{g\,\frac{\partial f}{\partial y}-f\,\frac{\partial g}{\partial y}}{g^{2}}, \qquad \frac{\partial}{\partial z}\Bigl(\frac{f}{g}\Bigr) =\frac{g\,\frac{\partial f}{\partial z}-f\,\frac{\partial g}{\partial z}}{g^{2}}.

Passo 4

Juntando as três componentes no vetor gradiente, temos
\nabla\Bigl(\frac{f}{g}\Bigr) =\frac{1}{g^{2}}\Bigl[ \bigl(g\,f_{x}-f\,g_{x}\bigr)\mathbf{i} +\bigl(g\,f_{y}-f\,g_{y}\bigr)\mathbf{j} +\bigl(g\,f_{z}-f\,g_{z}\bigr)\mathbf{k} \Bigr],
onde escrevemos f_{x}=\frac{\partial f}{\partial x}, etc. Reconhecendo
\nabla f=f_{x}\mathbf{i}+f_{y}\mathbf{j}+f_{z}\mathbf{k} e
\nabla g=g_{x}\mathbf{i}+g_{y}\mathbf{j}+g_{z}\mathbf{k}, chegamos a

Resposta

\nabla\Bigl(\frac{f}{g}\Bigr) =\frac{g\,\nabla f - f\,\nabla g}{g^{2}}.

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