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I know that

$$f_x=e^{t(x)}$$

(where the notation $f_x=\frac{df}{dx}$)

(EDIT: $f=f(x)$ and $t$ parameterizes $x$, so $x=x(t) \Leftrightarrow t=t(x)$)

and that therefore

$$\frac{d^n f_x}{dx^n}=\frac{e^t}{\dot{x}^n}$$

(where $\dot{x}=\frac{dx}{dt}$)

I need to find out what the function $f(x)$ is from this information. Help appreciated.

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    You don't know $t$ either?2012-08-08
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    $t$ parameterizes $x$. see edit above2012-08-08
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    "therefore" is wrong -- the second displayed equation doesn't follow from the first; this is only the case if $\dot x$ is constant.2012-08-08
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    $t=t(x)$ is not equivalent to $x=x(t)$. Please!2012-08-08
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    I didn't say they are equivalent. The double arrow means "implies." Doesn't $x=x(t)$ imply $t=t(x)$?2012-08-08
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    @joriki Could you spell that out a little more? As I see it, the nth derivative of the first equation is given by the second equation. I don't see how the question of whether or not $\dot{x}$ is constant is relevant.2012-08-08
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    @ben: Differentiating once yields $$ \frac{\mathrm df_x}{\mathrm dx}=\frac{\mathrm e^t}{\dot x}\;. $$ Then differentiating again yields $$ \frac{\mathrm d^2f_x}{\mathrm dx^2}=\frac{\mathrm e^t}{\dot x^2}-\frac{\mathrm e^t\ddot x}{\dot x^3}\;, $$ which is only equal to your expression if $\ddot x=0$.2012-08-08
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    Thanks for spelling that out. I am confused about one thing: it seems to me the second term in your second equation should be $$-\frac{e^t}{\dot{x}^2}\frac{d \ddot{x}}{dx}$$. You carried out the chain rule through differentiation w.r.t. $t$, but I don't see where you continued on to differentiation w.r.t. $x$? I was not sure what to make of $\frac{d\ddot{x}}{dx}$ and so assumed it was 0. If that is wrong please show me what it should be. Thanks.2012-08-08
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    Funny, I found this comment because I noticed that you just accepted an old answer of mine :-) You didn't ping me, so I might never have noticed it otherwise. About your question: I don't know how you got $\def\ma{\mathrm}\mathrm d\ddot x/\mathrm dx$ -- that's two differentiations more than $\dot x$ has (one more dot and one more $\ma d/\ma dx$), and we're only differentiating $\dot x$ once, so that can't be right. What I did was$$\frac{\mathrm d}{\ma dx}\dot x=\frac{\ma dt}{\mathrm dx}\frac{\mathrm d}{\ma dt}\dot x=\left(\frac{\ma dx}{\mathrm dt}\right)^{-1}\ddot x=\frac{\ddot x}{\dot x}\;.$$2012-08-09
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    I could have just left it as $\mathrm d\dot x/\mathrm dx$, but I figured that wasn't in the spirit of the game, since you'd turned $\mathrm dt/\mathrm dx$ into $1/\dot x$ in the question.2012-08-09
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    @joriki: I thought maybe everyone gave up on this one because of my 53% answer rate. Hence I went back and accepted a bunch of answers. I didn't realize you have to ping. As for the problem, ah yes I meant $d\dot{x}/dx$, not double dot. Besides, after thinking about it for two minutes I understood your derivative was correct. In the meantime I've figured out what $f$ is on my own, based on the assumption or postulate that $\frac{d\ln{x}}{d\ln{t}}=1$. Thanks for the help.2012-08-09
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    Well, you can integrate $\def\m{\mathrm d}\def\f#1#2{\frac{\m#1}{\m#2}}\m\ln x=\m\ln t$ to $\ln x=\ln t+C$, and thus $x=ct$, so $\dot x=c$ is constant, so that implies my condition.2012-08-10
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    @joriki: Ah yes indeed. Ok, here is my last try for this post, this time putting no conditions on $\dot{x}$: $$\frac{df_x}{dx}=\frac{e^t}{\dot{x}} \Rightarrow \frac{df}{dx}dx=e^t dt$$ Integrate both sides to get: $$f=e^t+C$$ Does that look legitimate?2012-08-15
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    @joriki: Nevermind, I don't know what I was thinking.2012-08-15

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