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Consider the derivative of y with respect to t:
dy/dt, according to the chain rule:

$\frac{dy}{dt} = (\frac{dy}{dx})(\frac{dx}{dt}) $

This means that the derivative of $y$ with respect to a variable $t$ is the coordinate of $y$ with respect to $x$(i.e. the slope at $x$) times the coordinate of $x$ with respect to $t$.

How would it make sense more intuitively?

Intuitively in the sense, that it would make more sense physically or well, intuitively rather than just analytically, i.e. just using algebraic laws, it would be rather meaningful, if it made sense in a sort of a logical argument.

And how do we take this further to the second order deriative of parametric equations:
d2y/dx2 = (dy'/dt)/(dx/dt)

bzal
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  • @SimplyBeautifulArt , the question is similar, but the answer is not what I was looking for this question, – bzal Mar 04 '17 at 15:23
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    This is not intuitive enough for you?$$\frac{dy}{dx}=\frac{dy}{\color{#4499de}{du}}\frac{\color{#4499de}{du}}{dx}$$ – Simply Beautiful Art Mar 04 '17 at 15:26
  • I object to the fact that two different functions are both called $y$ when we write the chain rule this way. On the right we have a function $y$, but on the left we have a different function that for the sake of clarity needs its own name, for example $\hat y$, The function $\hat y$ is defined by $\hat y(t) = y(x(t))$. This is a notational problem that afflicts many presentations of the chain rule. – littleO Mar 04 '17 at 15:32
  • I think this answer gives a very clear intuition for the chain rule: http://math.stackexchange.com/a/2084347/40119 – littleO Mar 04 '17 at 15:36
  • @SimplyBeautifulArt, it's analytically intuitive, (e.g. we can't relate things, may be I should had mentioned it's intuitive interpretation. – bzal Mar 04 '17 at 15:59
  • @littleO , please see my comment above, (it's the same link) – bzal Mar 04 '17 at 16:00
  • @bzal It's not even that rigorous man. How much more could you want than cancelling fractions?! – Simply Beautiful Art Mar 04 '17 at 16:01
  • @SimplyBeautifulArt , in modern era, I saw Euclid's GCD algorithm being proved using tools of abstract algebra, but during his time, he did it using only common logic(e.g. he said: "if we divide HG into FS equal parts we get AB which is equal to ......" and he PROVED his algorithm, and I found that making more sense. – bzal Mar 05 '17 at 01:51

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