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

by Denny   Last Updated January 19, 2018 04:20 AM

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

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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!

Tags : derivatives


Answers 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 : $$\frac{d}{dx} = \frac{dt}{dx}\frac{d}{dt}$$

David Reed
David Reed
January 19, 2018 04:16 AM

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