Derivatives With Multiple Branches

I have the derivatives of two functions, $$\frac{df(t)}{dt}$$ and $$\frac{dg(t)}{dt}$$. I would like to calculate the derivative $$\frac{df(t)}{dg(t)}$$.

This is a reasonably simple problem:

$$\begin{equation} \frac{df(t)}{dg(t)} = \frac{\frac{df(t)}{dt}}{\frac{dg(t)}{dt}} \end{equation}$$

However, suppose I only have access to a variable $$x$$ which defines $$t$$:

$$x = \cosh(t-0.5)$$, $$x\in\{0,1\}$$

The inverse equation requires two branches to describe properly:

$$t = \pm\cosh^{-1}(x)+0.5$$, $$t\in\{-0.5,0.5\}$$

So now I would like to calculate:

$$\begin{equation} \frac{df(t(x))}{dg(t(x))} = \frac{\frac{df(t(x))}{dt(x)}}{\frac{dg(t(x))}{dt(x)}} \end{equation}$$

I will surely have two solutions, for which there will be degenerate solutions at each value of $$x$$: one solution for each domain in $$t$$.

I have never dealt with such a situation before.

Is my understanding correct, and if so, what is the correct way of rendering $$\frac{df(t(x))}{dg(t(x))}$$? Is what I am attempting even a valid thing to do?

• What is with the absolute value? $$\frac{df(t)}{dg(t)} = \dfrac{\frac{df(t)}{dt}}{\frac{dg(t)}{dt}}$$ when the two right-side derivatives exist and the derivative of $g(t)$ is not $0$. Nov 26 '21 at 4:11

Assuming you know the derivatives of $$f(t),g(t)$$, the expression you wish to find is a function of $$t$$ too. $$F(t)=\frac{\,df(t)}{\,dg(t)}=\frac{f'(t)}{g'(t)}$$ The only use of $$x$$ is to find the value of $$t$$ at which you wish to find $$F(t)$$. And of course,for $$x=\cosh(t-0.5)$$, you get two valid values of $$t$$, hence a valid branch must be specified.