SOLUTION: sort of a calculus question... Find the derivative of the inverse function (f^-1)'(a) if f(x)=(e^x)+x and a=1. I've been stuck on this problem for over an hour now. I figured you s

Question 1034903: sort of a calculus question... Find the derivative of the inverse function (f^-1)'(a) if f(x)=(e^x)+x and a=1. I've been stuck on this problem for over an hour now. I figured you should find the inverse before the derivative but I'm having trouble with that. I've tried taking the natural logarithm of both sides. I can't get x by itself. Even when I try to use other logarithm rules. I also asked my classmates and they are stuck too.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

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