Rolle's theorem for a equation Let $f:R \to(0,+\infty)$ be a differtiable function, and $f(1)=1$. Prove that there is at least on $ξ \in (0,1)$ s.t. $$f(ξ)=\frac{1}{e^\frac{ξf'(ξ)}{f(ξ)}}$$
My try was to maybe use Rolle's theorem. Now $$e^\frac{xf'(x)}{f(x)}=\frac{1}{f(x)} $$ But I have no clue as on how to continue.
$$\frac{d}{dx}(\frac{xf'(x)}{f(x)})= \frac{f'(x)+xf''(x)+x(f^{'}(x))^2}{f^2(x)}$$ which is not that close to $\frac{1}{f(x)}$.
Can anyone help? 
 A: This is equivalent to proving there exists some $\xi$ such that $f(\xi) e^{\xi\frac{f'(\xi)}{f(\xi)}} = 1$. Since the range of $f$ is strictly positive, we may take a logarithm of it.
\begin{align}
f(x)e^{x \frac{f'(x)}{f(x)}} =& e^{\ln(f(x)) + x \frac{f'(x)}{f(x)}} \\
=& e^{\ln(f(x) + x \frac{d}{dx}\ln(f(x))}
\end{align}
So we have to show that this exponent equals $0$ somewhere in $(0,1)$. Define $g(x) := \ln(f(x))$. Then we know $g$ is a differentiable function such that $g(1) = 0$. This is exactly the $\xi$ you were looking for.
Let us look at the following integral.
\begin{align}
\int_0^1 g(x) + xg'(x) dx =& \int_0^1 g(x)dx + \int_0^1 x dg(x) \\
=& \int_0^1 g(x) dx + [xg(x)]_0^1 - \int_0^1 g(x)dx \\
=&0
\end{align}
 Hence, by Rolle's theorem on the function $t \mapsto \int_0^t g(x)+xg'(x)dx$, there must be some value $\xi \in (0,1)$ such that $g(\xi) + \xi g'(\xi) = 0$.
A: Thanks to @Lukas Rollier for the help, I managed to solve it in a different way,(using his help),probably a bit easier.For the record:
Using $$f(x)e^{x\frac{f'(x)}{f(x)}}=1 \Rightarrow e^{ln(f(x))(x)'+x(ln(f(x))'}=1 \Rightarrow (xln(f(x))'=0 $$
And if we set $t(x)=xln(f(x))$, we have $t(0)=0, t(1)=0$ and apply Rolle's theorem.
