Question 1034903
the derivative of e^x + x is e^x + 1.
i'm not that smart.
i used an online derivative calculator to find it.


you get y = e^x + 1 is the equation of the derivative of y = e^x + x.


the inverse y = e^x + 1 is x = e^y + 1.
you interchange the variables.
x = y and y = x
that's your inverse equation.


subtract 1 from both sides of that equation to get x - 1 = e^y
take the natural log of both sides of that equation to get ln(x-1) = ln(e^y)
since ln(e^y) = y*ln(e), and since ln(e) = 1, the equation becomes:
ln(x-1) = y
this can be shown as y = ln(x-1).


y = ln(x-1) is equivalent to x = e^y + 1 and is therefore the derivative equation of y = e^x + x, as far as i can tell.


what i end up with is that the inverse equation of the derivative of y = e^x + x winds up being y = ln(x-1).


if you're saying that f(x) = ln(x-1) is the derivative equation, and you want to find f(a), then the equation becomes f(a) = ln(a-1).


if a = 1, then the equation becomes f(1) = ln(1-1) = ln(0) which is undefined.


i'm not totally sure i did the right thing, but it kind of makes sense to me.
don't take it as gospel though, because i'm not totally sure it's what you want.
it's the best i can do however, so take it for what it's worth.