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I saw this in Quora: $$ \frac{d}{dt}\, \frac{dy}{dx} = \left(\frac{d}{dx}\, \frac{dy}{dx}\right) \frac{dx}{dt}. $$

I tried understanding this equation but I couldn't. I think it's not valid after searching wikipedia and multiple text books.

I know that there is composition going on but I can't really figure out what is a function in this expression.

I understand all the other notations but this one boggles me.

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    $\begingroup$ I think that a possibly better way to write this would be $$\frac{\mathrm{d}}{\mathrm{d}t}y'(x(t)) = y''(x(t))x'(t)$$ $\endgroup$ – Michael Lee Aug 3 '17 at 10:21
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    $\begingroup$ This is the chain rule applied to ${dy\over dx}$. $\endgroup$ – Edu Aug 3 '17 at 10:21
  • $\begingroup$ @MichaelLee Oh ok, that makes sense. I thought it was the definition of the chain rule but it's actually the chain rule applied on a derivative. Right ? $\endgroup$ – 3366784 Aug 3 '17 at 10:29
  • $\begingroup$ @3366784: I made a couple of minor edits, changing "definition" to "application" in the title and "equation" in the body, and LaTeX-ifying your equation. Please feel free to edit or roll back if that doesn't match your intent. $\endgroup$ – Andrew D. Hwang Aug 3 '17 at 11:46
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We can write the definition of the chain rule as $$\frac{dz}{dt} = \frac{dz}{dx} \cdot \frac{dx}{dt}.$$ Now, take $z={dy\over dx}$ and plug it in the previous formula: $${d\over dt}{dy\over dx}=\left({d\over dx}{dy\over dx}\right){dx\over dt}\quad\text{(1)}$$ You get the expression of your question.

In particular, $\text{(1)}$ is not a definition of the chain rule, but an application of it.

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    $\begingroup$ Clarification: the last phrase in this answer referred to the question before its edition, which asked about $\text{(1)}$ being a correct definition of the chain rule. $\endgroup$ – Edu Aug 3 '17 at 11:56

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