# Thread: Derivative confusion

1. ## Derivative confusion

Hello everyone!

I've lately came accross this "confusion" as I was solving a cauchy-euler equation that required a subtitution of $\displaystyle e^t$ for $\displaystyle x$. It followed that $\displaystyle x=e^t\Leftrightarrow t=ln|x|.$ Then $\displaystyle dt=\frac{1}{x}\,dx$ or simply $\displaystyle \frac{dt}{dx}=\frac{1}{x}$. But what about $\displaystyle \frac{dx}{dt}$? Is it $\displaystyle x$ or $\displaystyle e^t$. I'm very confused about this because whenever we have a 1st order ODE let's say $\displaystyle y'=f(x)$ we change it to $\displaystyle \frac{dy}{dx}=f(x)$ and then multiply the RHS by $\displaystyle dx$ follows that $\displaystyle dy=f(x)\,dx$ as if the latter were a denominator!

If someone clears that out for me I would be very greatful!
Thanks!!

2. Originally Posted by rebghb
Hello everyone!

I've lately came accross this "confusion" as I was solving a cauchy-euler equation that required a subtitution of $\displaystyle e^t$ for $\displaystyle x$. It followed that $\displaystyle x=e^t\Leftrightarrow t=ln|x|.$ Then $\displaystyle dt=\frac{1}{x}\,dx$ or simply $\displaystyle \frac{dt}{dx}=\frac{1}{x}$. But what about $\displaystyle \frac{dx}{dt}$? Is it $\displaystyle x$ or $\displaystyle e^t$. I'm very confused about this because whenever we have a 1st order ODE let's say $\displaystyle y'=f(x)$ we change it to $\displaystyle \frac{dy}{dx}=f(x)$ and then multiply the RHS by $\displaystyle dx$ follows that $\displaystyle dy=f(x)\,dx$ as if the latter were a denominator!

If someone clears that out for me I would be very greatful!
Thanks!!
Hi rebghb,

$\displaystyle since\ x=e^t$

$\displaystyle \frac{dx}{dt}=e^t=x=\frac{1}{\left[\frac{dt}{dx}\right]}$

As $\displaystyle \frac{dy}{dx}$ is a ratio at any given x, being the slope of the tangent to the graph, you may treat $\displaystyle dx$ as a denominator,

whereby integration is performed as anti-differentiation.

$\displaystyle y\ '(x)=f(x)$

$\displaystyle \frac{dy}{dx}=f(x)$

$\displaystyle dy=f(x)dx$

$\displaystyle \int{dy}=y=\int{f(x)}dx$

3. [quote=Archie Meade;498895]Hi rebghb,

$\displaystyle since\ x=e^t$

$\displaystyle \frac{dx}{dt}=e^t=x=\frac{1}{\left[\frac{dt}{dx}\right]}$

As $\displaystyle \frac{dy}{dx}$ is a ratio at any given x, being the slope of the tangent to the graph, you may treat $\displaystyle dx$ as a denominator,
quote]

Thanks, well I am aware that $\displaystyle \frac{dy}{dx}$ is essentially a slope, but when we can treat $\displaystyle dx$ as a denominator, can we treat $\displaystyle dy$ as a numerator? (if the bottom part is the denominator then what about the upper part). I got your explanation using the definition of an integral (well not exactly definition) but if $\displaystyle x=e^t\frac{dx}{dt}=e^t$ but what about $\displaystyle \frac{dt}{dx}$?? Why isn't it just the inverse as in $\displaystyle frac{1}{x}$. I'm lost because noone has made this clear, deriving has always been a routine

4. Yes. While $\displaystyle \frac{dy}{dx}$ is NOT a fraction, it is a limit of one and so we can typically treat it like a fraction. That is one of the reasons for defining the "differential", dx and dy with dy= f'(x)dx. That allows us to formally treat dy/dx as a fraction.

5. Okay I'll ask this bluntly, this will sound stupid: Why is it thay $\displaystyle \frac{dx}{dt}=e^t$ and $\displaystyle \frac{dt}{dx}=\frac{1}{x}$ and not $\displaystyle e^{-t}$ ??

6. It is exp(-t).

$\displaystyle \frac{dx}{dt}=e^t=x$

$\displaystyle \frac{dt}{dx}=\frac{1}{x}=\frac{1}{e^t}=e^{-t}=\frac{1}{\frac{dx}{dt}}$

7. aaaah! Thanks!