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Can somebody please check my working? The variable $y$ depends on $x$ and $x$ and $t$ are related by $x=e^t$

Show that $x\frac{dy}{dx}=\frac{dy}{dt}$

$$x=e^t$$ $$dx=e^t dt$$ $$\frac{dx}{dt}=e^t=e^t\times1=e^t \times\frac{dy}{dy}$$ $$\frac{dx}{dt}\times\frac{dy}{dx}=x\frac{dy}{dy}\times\frac{dy}{dx}$$ $$\frac{dy}{dt}=x\frac{dy}{dx}$$

Is this correct?

mathnoob123
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    You treat $\frac{dy}{dx}$ as a fraction... A mathematician can go crazy for this. :) – Ixion May 22 '17 at 21:54
  • Yeah. Same. Dealing dy/dx as fractions can make maths go haywire. – mathnoob123 May 22 '17 at 21:57
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    @Ixion: There is no problem with treating the differentials as fractions. The notation was adopted as it allows to represent the chain rule very intuitively. –  May 22 '17 at 22:05
  • Your development is dubious. There is no connection between the last two lines. –  May 22 '17 at 22:06
  • @YvesDaoust I think you're right, but formally a student like me may lose the real meaning of what he's doing – Ixion May 22 '17 at 22:09
  • @Ixion: I don't see how. Actually, in modern notation $dx$ and $dy$ are differentials, and it is rigourous to treat them as factors. The equations $dy=f'(x),dx$ and $dy/dx=f'(x)$... are perfectly valid. –  May 22 '17 at 22:12
  • @YvesDaoust: actually, I really don't know what to say. I'm not a mathematician (yet). :) – Ixion May 22 '17 at 22:18

2 Answers2

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In your answer, I am not sure how to get from line (3) to line (4).

I would proceed using the Chain Rule: $$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} $$ but you know how $x$ depends on $t$, can you find $dx/dt$ and plug in?

gt6989b
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  • Well I wrote $=e^t1=e^t\frac{dy}{dy}$. Is it incorrect? – mathnoob123 May 22 '17 at 21:52
  • @FaiqRaees I don't understand your comment. $dy/dy = 1$ so $e^t = e^t \frac{dy}{dy}$, but how does this help anyone? You already have all you need in terms of $y$, you just want to compute $dx/dt$ in terms of $x$ itself, not in terms of $y$. – gt6989b May 22 '17 at 21:54
  • @FaiqRaees I get it -- you think $dy/dy$ is a fraction? This is certainly wrong then (although it does behave somewhat similarly, but you cannot apply standard fraction arithmetic rules to it). – gt6989b May 22 '17 at 21:55
  • And for line 3-4 I multiplied dy/dx on both sides. Is it incorrect? – mathnoob123 May 22 '17 at 21:56
  • Oh. I do understand it is just dy/dy is just change in y/ change in y, but I thought it could behave as fractions like for integrations when we are making a substitution we deal these derivatives as functions. Why is that correct over there? – mathnoob123 May 22 '17 at 21:58
  • @gt6989b Such methods are commonly taught to applied students and work perfectly well here. In the context of this question, differentials are fractions. – Kaynex May 22 '17 at 21:59
  • @Kaynex I wish people who teach such fantasy were disqualified from teaching until they confess and repent. No, differentials are never fractions and it is quite tricky to know when they do, and when they don't behave as ones. There is a numerous amount of places where such "techniques" will create a nightmare in your calculations. – gt6989b May 22 '17 at 22:02
  • @Kaynex you're right, but I can tell you my teacher would become crazy if I wrote something like that. Anyway I think Leibniz would agree with Faiq Reaees. – Ixion May 22 '17 at 22:03
  • Btw I am a pre-university student so I haven't been taught calculus at a very high level. So maybe we were introduced these simplicities to make the comprehension of the topic easier. – mathnoob123 May 22 '17 at 22:13
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    You will find this very instructive then -- https://math.stackexchange.com/questions/1906241/when-not-to-treat-dy-dx-as-a-fraction-in-single-variable-calculus – gt6989b May 22 '17 at 22:16
  • And then there's "non-standard analysis" which uses the Leibniz notion of infinitesimals: here – Χpẘ May 22 '17 at 23:28
  • @Χpẘ yes, there is, but that falls way out of the level of appreciation of the OP – gt6989b May 23 '17 at 02:53
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You are given $$y=f(x),x=e^t.$$

Then

$$\frac{dy}{dx}=f'(x)$$ and by the chain rule,

$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=f'(x)e^t=f'(x)x,$$ hence the claim.