Solve $\frac{dy}{dx} = f^{-1}(x)$ I am doing a differential equations subject, and have been given this as a challenge question.
Solve $\frac{dy}{dx} = f^{-1}(x)$
($f^{-1}(x)$ is the inverse function of x.)
I assume that the answer involves separable differential equations, but I can't quite see how to do it.
Is there a substitution that will help solve this?
Edit:
I've tried the following ideas:
Idea 1.
$$
\begin{align}
y &= \int f^{-1}(x) \, dx\\
\mathrm{let}\,\,u &= f^{-1}(x)\\
x &= f(u)\\
\frac{dx}{du} &= f'(u)\\
dx &= f'(u) \, \, du\\
\mathrm{so} \,\, y &= \int u f'(u)\, du\\
\end{align}
$$
Is this right, and if so, would I just integrate by parts?
 A: I misunderstood earlier - sorry.
FWIW, there is no such function defined on all of $\mathbb{R}$.
If there were, by referring to the derivative at all I would assume that we would be looking for a continuous function (say as opposed to one continuous almost everywhere). And by referring to an inverse I would assume that we would be looking for a one-to-one function (as opposed to a function with a restricted inverse.)
Since the domain of $f$ would be all of $\mathbb{R}$, the range of $f^{-1}$ is all of $\mathbb{R}$ too. Then since $f'=f^{-1}$, $f$ is sometimes increasing, sometimes decreasing. This is not possible for a one-to-one continuous function.
A: Look at this Hint.

Theorem: Let $f$ be a one-to-one function with inverse $f^{-1}$. Suppose $f$ be differentiable at a point $x$, with $f'(x)\ne0$ and moreover suppose $f^{-1}$ is continuous at $f(x)=y$. Then $$(f^{-1})'=\frac{1}{f'(x)}$$

In fact, $$\frac{dx}{dy}\cdot f'(x)=1$$ Now think of this ODE instead, $$y'=f^{-1}\longrightarrow x'\cdot x=1$$
