Show that $\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot \frac{d\vec{r}}{dt}$ where F is a differentiable function of  $x, y,z, t$ and $x, y, z$ differentiable functions of $t$, Show that $\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot \frac{d\vec{r}}{dt}$
We define $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ as the position vector
I would like you to give me some suggestions. It has been difficult for me to know where to start
I did this but I don't know if it's okay
$\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$
$\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$
$\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$
 A: Write the multivariate chain rule in the language of infinitesimals, viz.$$dF=\frac{\partial F}{\partial t}dt+\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz.$$Note there is nothing different about $t$ here as compared with $x,\,y,\,z$. But if $x,\,y,\,z$ are differentiable functions of $t$ so an $F(t,\,x,\,y,\,z)$ differentiable with respect to each of its four arguments simplifies to a differentiable function of $t$, division by $dt$ gives$$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}.$$Finally, write the last three terms as a dot product:$$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot\frac{d\vec{r}}{dt}.$$
A: Use the chain rule to get, $$\dfrac{dF}{dt}=\dfrac{\partial F}{\partial t}+\dfrac{\partial F}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial F}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial F}{\partial z}\dfrac{dz}{dt}$$
Should be clear from there.
