How to differentiate $O=f(x,t)g(x,t)h(x,t)$ with respect to t? For
$$O=f(x,t)g(x,t)h(x,t),
$$
what is $dO \over dt$? 
$O$ is a function of $f,g,h$ which are in turn functions of $x,t$. $x$ is not dependent on $t$.
I read that the first two steps to the solution are
$$\frac{dO}{dt}=\frac{dO}{df} \frac{df}{dt} + \frac{dO}{dg} \frac{dg}{dt} + \frac{dO}{dh} \frac{dh}{dt}=gh\frac{df}{dt}+fh\frac{dg}{dt}+fg\frac{dh}{dt}$$
I have problems understanding these first two steps.
What is the differentiation rule used for the first step? If I cancel the pairs of $df$, $dg$ and $dh$, I get 
$$\frac{dO}{dt}=\frac{dO}{dt}+\frac{dO}{dt}+\frac{dO}{dt}.$$
That doesn't seem to make sennse. 
Also, for the second step, what differentiation rule was used to get $$\frac{dO}{df}=gh, \frac{dO}{dg}=fh,\frac{dO}{dh}=fg$$?
Edit: There was some discussion that total time derivatives should not be used. However, the full solution given was 
$$\frac{dO}{dt}=\frac{dO}{df} \frac{df}{dt} + \frac{dO}{dg} \frac{dg}{dt} + \frac{dO}{dh} \frac{dh}{dt}$$$$=gh\frac{df}{dt}+fh\frac{dg}{dt}+fg\frac{dh}{dt}$$$$=gh(\frac{\partial f}{\partial t} \frac{dt}{dt} + \frac{\partial f}{\partial x} \frac{dx}{dt}) + fh(\frac{\partial g}{\partial t} \frac{dt}{dt}+ \frac{\partial g}{\partial x} \frac{dx}{dt}) + fg(\frac{\partial h}{\partial t} \frac{dt}{dt} + \frac{\partial h}{\partial x} \frac{dx}{dt}) $$ $$=gh \frac{\partial f}{\partial t} + fh \frac{\partial g}{\partial t} + fg\frac{\partial h}{\partial t}$$
 A: You can see the problem in two ways.
1. Functions composition
Naming $o : (f,g,h) \mapsto fgh$, you have 
$$O(x,t) = o(f(x,t), g(x,t), h(x,t))$$ based on that, you can use chain rule to say
$$\frac{\partial O}{\partial t}=\frac{\partial f}{\partial t}\frac{\partial o}{\partial f}+\frac{\partial g}{\partial t}\frac{\partial o}{\partial g}+\frac{\partial h}{\partial t}\frac{\partial o}{\partial h}.$$
We also have
$$\begin{cases}
\frac{\partial o}{\partial f} &= gh\\
\frac{\partial o}{\partial g} &= fh\\
\frac{\partial o}{\partial h} &= fg
\end{cases}$$
Leading finally to $$\frac{\partial O}{\partial t}=gh\frac{\partial f}{\partial t}+fh\frac{\partial g}{\partial t}+fg\frac{\partial h}{\partial t}.$$
this is essentially the solution you mention in your question. Now what I don't like in the provided solution is that $o$ and $O$ are not clearly separated which is confusing.
2. Product of functions
There is another way to deal with the problem, writing $O = (fg)h$ and using twice the differentiation of the product of two functions.
A: It seems like there was a typo by the writer. 
The correct first two steps to the solution should be 
$$\frac{dO}{dt}=\frac{\partial O}{\partial f} \frac{df}{dt} + \frac{\partial O}{\partial g} \frac{dg}{dt} + \frac{\partial O}{\partial h} \frac{dh}{dt}$$$$=gh\frac{df}{dt}+fh\frac{dg}{dt}+fg\frac{dh}{dt}$$
