Is there a function that it's derivative is its inverse?

Question

kainui · Answer

That doesn't answer my question.

miracrown · Answer

Right of the bat, i honestly do not know...
My intuition is that there probably is a small class of functions where the derivative is also the inverse function

miracrown · Answer

Did you do a search on this topic at all?

dumbcow · Answer

If you can find a function where its product with its derivative is 1 that will qualify it as having inverse = derivative http://en.wikipedia.org/wiki/Inverse_function#Inverses_and_derivatives

miracrown · Answer

hmm.. that is a bit surprising... considering how almost every question conceivable is dealt with by someone in cyberspace!

I was researching and finding general info re: inverse functions
need to refine the search

miracrown · Answer

I am curious now as well --- to see if maybe due to the nature of the equation between the deriv and the inverse and the deriv of the inverse, it may not be possible...

miracrown · Answer

What about a constant function or 0 ?
	f(x) = 0
	f^-1(x) = 0
	f'(x) = 0

geerky42 · Answer

If $f(x)=0$, then $f^{-1}(x)$ doesn't exist.

miracrown · Answer

I guess any other constant would not satisfy the condition
as an inverse is reflected around the line y=x

Just might be a function that is identically 0 is the only one
just like e^x is only function where f(x) = f'(x)

miracrown · Answer

ahh :o  @geerky42 - I thought that might be the case ..

miracrown · Answer

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