How to prove this identity involving vector calculus.

Question

I have the following functions $$\phi_1=\mathbf{F}\cdot \frac{\partial \mathbf{r}}{\partial u}$$ and $$\phi_2=\mathbf{F}\cdot \frac{\partial \mathbf{r}}{\partial v}$$ where $\mathbf{F}$ is a vector field in $\mathbf{r}$ is the point $(x,y,z)$ is a point in space. Then I want to show that $$\frac{\partial \mathbf{\phi_1}}{\partial v}-\frac{\partial \mathbf{\phi_2}}{\partial u}=(\nabla \phi_1)\cdot \frac{\partial \mathbf{r}}{\partial v}-(\nabla \phi_2)\cdot \frac{\partial \mathbf{r}}{\partial u}.$$ I computed the first term $$\frac{\partial \mathbf{\phi_1}}{\partial v}-\frac{\partial \mathbf{\phi_2}}{\partial u}=\frac{\partial \mathbf{F}}{\partial v}\cdot \frac{\partial \mathbf{r}}{\partial u}-\frac{\partial \mathbf{F}}{\partial u}\cdot \frac{\partial \mathbf{r}}{\partial v}.$$ Then I computed $$(\nabla \phi_1)\cdot \frac{\partial \mathbf{r}}{\partial v}-(\nabla \phi_2)\cdot \frac{\partial \mathbf{r}}{\partial u}= \frac{\partial \mathbf{F}}{\partial u}\cdot \frac{\partial \mathbf{r}}{\partial v}-\frac{\partial \mathbf{F}}{\partial v}\cdot \frac{\partial \mathbf{r}}{\partial u}$$ where I did $$\nabla \phi_1\cdot \frac{\partial \mathbf{r}}{\partial v}=\frac{\partial \phi_1}{\partial \mathbf{r}}\cdot \frac{\partial \mathbf{r}}{\partial v}=\frac{\partial \mathbf{F}}{\partial u}\cdot \frac{\partial \mathbf{r}}{\partial v }+\mathbf{F}\cdot \frac{\partial^2}{\partial u \partial v}$$ and $$\nabla \phi_2\cdot \frac{\partial \mathbf{r}}{\partial u}=\frac{\partial \phi_2}{\partial \mathbf{r}}\cdot \frac{\partial \mathbf{r}}{\partial u}=\frac{\partial \mathbf{F}}{\partial v}\cdot \frac{\partial \mathbf{r}}{\partial u }+\mathbf{F}\cdot \frac{\partial^2}{\partial v \partial u}.$$ As you can probably see that the two expressions are not the same, there is a minus sign which is coming up. How do I fix this?

Answer 1

Hint: The gradient of an inner product in $\mathbb{R}^3$ is \begin{equation} \nabla(\textbf{A}\cdot\textbf{B}) = (\textbf{A}\cdot\nabla)\textbf{B}+(\textbf{B}\cdot\nabla)\textbf{A}+\textbf{A}\times(\nabla\times\textbf{B})+\textbf{B}\times(\nabla\times\textbf{A}) \end{equation}