Prove this vector identity Let $f$ and $g$: $\mathbb R^3 \rightarrow \mathbb R$ be $C^{1}$ scalar functions. Prove that
$$ \nabla \left( \frac{f}{g} \right) = \frac{1}{g^2}\left( g\nabla f - f\nabla g \right)$$ $$g \neq 0$$  
 A: The $i$-th component of the gradient is:
$$
\frac{\partial }{\partial x_i} \left(\frac{f}{g}\right) = \left(\frac{\partial f}{\partial x_i} g - \frac{\partial g}{\partial x_i} f\right)\frac{1}{g^2},
$$
thus
\begin{align}\nabla \left(\frac{f}{g}\right) &= \left(\frac{\partial }{\partial x_1} \left(\frac{f}{g}\right), \frac{\partial }{\partial x_2} \left(\frac{f}{g}\right), \frac{\partial }{\partial x_3} \left(\frac{f}{g}\right)\right)
\\
&= \frac{1}{g^2} \left(\frac{\partial f}{\partial x_1} g - \frac{\partial g}{\partial x_1} f , \frac{\partial f}{\partial x_2} g - \frac{\partial g}{\partial x_2} f, \frac{\partial f}{\partial x_3} g - \frac{\partial g}{\partial x_3} f\right)
\\
&=\frac{1}{g^2} \left( \left(\frac{\partial f}{\partial x_1} g,\frac{\partial f}{\partial x_2} g, \frac{\partial f}{\partial x_3} g\right) - \left(\frac{\partial g}{\partial x_1} f,\frac{\partial g}{\partial x_2} f, \frac{\partial g}{\partial x_3} f\right)\right)
\\
& =  \frac{1}{g^2}\left(g\nabla f - f\nabla g\right),
\end{align}
and I believe there should be a parenthesis in your expression.
A: I will assume $f, g : \mathbb{R}^3 \to \mathbb{R}$.
We have 
$$\nabla\bigg(\frac{f}{g}\bigg) = \bigg\langle \frac{\partial f/g}{\partial x},\frac{\partial f/g}{\partial y}, \frac{\partial f/g}{\partial z} \bigg \rangle $$
By the quotient rule, the above becomes
$$\bigg \langle \bigg(\frac{\partial f}{\partial x} g - \frac{\partial g}{\partial x}f\bigg)\frac{1}{g^2},\bigg(\frac{\partial f}{\partial y} g - \frac{\partial g}{\partial y}f\bigg)\frac{1}{g^2},\bigg(\frac{\partial f}{\partial z} g - \frac{\partial g}{\partial z}f\bigg)\frac{1}{g^2} \bigg \rangle $$
which is exactly
$$\frac{1}{g^2}(g\nabla f - f\nabla g)$$  
