Inverse of the derivative for f(x) = f'(x)

Question

I'm new so forgive my inexperience here.

The problem concerns the following:

$$ f: \Bbb R \to (0, \infty), f(x) = f'(x) $$

The first part of the problem involves showing f is increasing, this seemed simple since:

$$ f(x) > 0 => f'(x) > 0$$

Therefore $f$ is increasing.

The next part of the problem is what confuses me, I need to show that $(f^{-1})'(y) = \frac{1}{y} $ , that is to say the derivative of the inverse of $f$ is $1 \over y$

I can see how the problem makes sense, i.e. a function that satisfies $f(x) = f'(x)$ is $e^x $, the inverse of which is $\ln(x)$ and the derivative of that is $\frac{1}{x}$. However simply stating that $e^x$ is a function for$ f(x)$ doesn't seem like a worthwhile attempt at a solution as there isn't a unique solution for $f(x) = f'(x)$ .

Then I thought that$ f'(x) = f(x) $can be solved as a simple differential equation, but then that solution doesn't seem suitable for the course material (basic Analysis).

Is there any other solution to the second part of the problem I am overlooking?

Thanks in advance.

Answer 1

that's a really nice problem, so I'm gonna post this solution anyway even though it doesn't follow the approach you proposed - simply consider $\frac{f(x)}{e^x}$ and calculate its derivative

Answer 2

Because $f^{-1}$ is the inverse of $f$, for all $y$, we have

$$ f(f^{-1}(y)) = y $$ and so, by the chain rule $$ \frac{d}{dy}f(f^{-1}(y)) = 1 \\ f'(f^{-1}(y)) \left(\frac{d}{dy}f^{-1}(y)\right) = 1 $$ and because $f' = f$, $$ f(f^{-1}(y)) \left(\frac{d}{dy}f^{-1}(y)\right) = 1 \\ y \left(\frac{d}{dy}f^{-1}(y)\right) = 1 \\ $$