1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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 $0$. $\endgroup$ Nov 26 '21 at 4:11
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.