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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? –  J. M. Aug 8 '12 at 5:18
    
$t$ parameterizes $x$. see edit above –  ben Aug 8 '12 at 5:24
    
"therefore" is wrong -- the second displayed equation doesn't follow from the first; this is only the case if $\dot x$ is constant. –  joriki Aug 8 '12 at 5:34
1  
@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 $\ddot x=0$. –  joriki Aug 8 '12 at 6:08
1  
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$$\frac{\mathrm d}{\ma dx}\dot x=\frac{\ma dt}{\mathrm dx}\frac{\mathrm d}{\ma dt}\dot x=\left(\frac{\ma dx}{\mathrm dt}\right)^{-1}\ddot x=\frac{\ddot x}{\dot x}\;.$$ –  joriki Aug 9 '12 at 23:23

1 Answer 1

My idea, excecution heavily dependend on $t(x)$:

$\ \ \ \ \frac{\text df(x)}{\text dx}=\text{e}^{t(x)}=\sum_{n=0}^\infty \frac{1}{n!}t(x)^n$

$\ \ \ \Longrightarrow\ \ \ f(x)=\sum_{n=0}^\infty \frac{1}{n!}\int t(x)^n \text d x,$

$\ \ \ \ \int t(x)^n \text d x\ \ \ \ ?$

Ansatz $t(x)^ng(x)$:

$\ \ \ \ \frac{\text d}{\text dx}(t^n g) =n t^{n-1}t'g+t^ng' =t^n(ngt'/t+g') =t^n(g\ \text{log}(t^n)'+g') \overset{!}{=}t^n$

To solve

$\ \ \ \ g'=1-\text{log}(t^n)'\ g,$

which will probably look something like $h_1(x)+h_2(x) \text e^{\int\ h_3(x) g(x)\ \text d x}$.

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