How to go from $dx = -t^2dt$ to $\frac{d}{dx} = -t^2\frac{d}{dt}$

Question

Suppose $x = 1/t$. So now $x$ is a function of $t$, i.e., $x(t)$.

So $$\frac{dx(t)}{dt} = -t^{-2} \Rightarrow dx(t) = -t^{-2}dt$$

This problem is from the textbook: advanced mathematical methods for scientists and engineers

How to go from "$dx = -t^2dt$" to "$\frac{d}{dx} = -t^2\frac{d}{dt}$"?

It seems that I just divide the previous term by $1$ and then multiply it by $d$.

However, it seems unrealistic to me; can anyone please explain this carefully to me? More specifically, can I multiply $d$, which is like an operator to me. thanks!

Answer 1

You use the inverse-function theorem and the chain rule:

Inverse function theorem says $$\frac{dx}{dt} = -t^{-2} \to \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}= -t^2 $$

The chain rule says : $$\frac{d}{dx} = \frac{dt}{dx}\frac{d}{dt} = -t^2\frac{d}{dt}$$

Answer 2

Correct me if wrong :

$x=x(t)$, differentiable on an interval $I$, and $x'(t)\not=0,$

then the inverse function $t=t(x)$ exists, and is differentiable $x=x(t).$

Let $F(t(x))$ arbitrary, differentiable, then:

$\dfrac{d}{dx} F(t(x)) = \dfrac{d}{dt} F(t)×\dfrac{d}{dx}t(x),$ I.e.,

$[\dfrac{d}{dx}]F(t(x)) = [\dfrac{d}{dx}t(x) \dfrac{d}{dt}]F(t(x))$,

or $\dfrac{d}{dx} = t'(x)\dfrac{d}{dt}$, as an operator equation.

With:

$t'(x) = \dfrac{1}{x'(t)}= \dfrac{1}{-t^{-2}}$:

$\dfrac{d}{dx} = -t^2 \dfrac{d}{dt}$.