How to differentiate $O=f(x,t)g(x,t)h(x,t)$ with respect to t?

Question

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 $$\frac{dO}{dt}=\frac{dO}{df} \frac{df}{dt} + \frac{dO}{dg} \frac{dg}{dt} + \frac{dO}{dh} \frac{dh}{dt}$$$$=gh\frac{df}{dt}+fh\frac{dg}{dt}+fg\frac{dh}{dt}$$$$=gh(\frac{\partial f}{\partial t} \frac{dt}{dt} + \frac{\partial f}{\partial x} \frac{dx}{dt}) + fh(\frac{\partial g}{\partial t} \frac{dt}{dt}+ \frac{\partial g}{\partial x} \frac{dx}{dt}) + fg(\frac{\partial h}{\partial t} \frac{dt}{dt} + \frac{\partial h}{\partial x} \frac{dx}{dt}) $$ $$=gh \frac{\partial f}{\partial t} + fh \frac{\partial g}{\partial t} + fg\frac{\partial h}{\partial t}$$

Answer 1

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.

Answer 2

It seems like there was a typo by the writer.

The correct first two steps to the solution should be $$\frac{dO}{dt}=\frac{\partial O}{\partial f} \frac{df}{dt} + \frac{\partial O}{\partial g} \frac{dg}{dt} + \frac{\partial O}{\partial h} \frac{dh}{dt}$$$$=gh\frac{df}{dt}+fh\frac{dg}{dt}+fg\frac{dh}{dt}$$