Prove this vector identity

Question

Let $f$ and $g$: $\mathbb R^3 \rightarrow \mathbb R$ be $C^{1}$ scalar functions. Prove that

$$ \nabla \left( \frac{f}{g} \right) = \frac{1}{g^2}\left( g\nabla f - f\nabla g \right)$$ $$g \neq 0$$

Answer 1

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

$$\frac{1}{g^2}(g\nabla f - f\nabla g)$$

Answer 2

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.