Differentiate inverse (e^x + ln x )

[Ambidext]

Homework Statement

Let f(x) = e^x + ln x

Find (f^-1) ' (e)

Homework Equations

let y = f^-1 x

The Attempt at a Solution

I tried finding the inverse of f(x) but got stuck. I arrived at:

x = e^y + ln y

How do I make y the subject of formula?

 

[rock.freak667]

You will not be able to algebraically find f^-1(x) using normal means. You may need to use a special function to help you.

 

[JonF]

use this rule: (f^-1)’(b) = 1/f’(a) where b=f(a).

To get a solve: e = e^a + ln(a)

 

[D H]

I tried finding the inverse of f(x) but got stuck. I arrived at:

x = e^y + ln y

How do I make y the subject of formula?

You appear to have a misunderstanding of the concept of an inverse function. The inverse f^-1 of some function f(x), if it exists, is the function that solves f^-1(f(x))=x. If you denote y=f(x) you can compute df^-1(y)/dy by the chain rule.

Try that.

Another aspect of the problem involves finding f^-1(e). In general you will need to use numerical methods to find f^-1(y). You do not need to resort to numerical methods to find f^-1(e). This has an exact and rather simple answer.

 

[Ambidext]

Thank you all, for the replies.

You appear to have a misunderstanding of the concept of an inverse function. The inverse f^-1 of some function f(x), if it exists, is the function that solves f^-1(f(x))=x. If you denote y=f(x) you can compute df^-1(y)/dy by the chain rule.

Try that.

Another aspect of the problem involves finding f^-1(e). In general you will need to use numerical methods to find f^-1(y). You do not need to resort to numerical methods to find f^-1(e). This has an exact and rather simple answer.

The question already gave y = f(x). Was simply lazy to use another letter for f^-1(x) But yeah, I do know that f^-1(f(x)) = x. But your method to get df^-1(x) / dx looks pretty good. Thanks!

 

[Ambidext]

By the way, I just realized my question was misleading the way I typed. I meant to differentiate f^-1(e), where f(x) = e^x + ln x, and not differentiate 1/f(e).

 

[D H]

There is admittedly a bit of ambiguity in whether f^-1 means the inverse function or the multiplicative inverse of the function. Nonetheless, we did figure out what you meant.

You still do want to differentiate f(x), by the way.

 

[Ambidext]

Well, I got:

f(e) = e^e + ln e

let z = f^-1(e)

e = e^z + ln z
ln e = z + ln (ln z)

z = ln e - ln y

dz/de = 1/e - 1/y (dz/de)
dz/de(1 - 1/y) = 1/e
dz/de = 1 / e(1 - 1/y)


ARGHHH AND I GOT STUCK!

 

[D H]

Well, I got:

f(e) = e^e + ln e

You should not be looking for f(e). You want some value x such that f(x)=e. In short, you want f^-1(e).

[QUOE]let z = f^-1(e)[/QUOTE]
You don't want to do this, either. This is defining z to be the specific number f^-1(e).

dz/de

This makes zero sense. z is a specific number here, not a variable, so the derivative with respect to some variable is zero. However, e is not a variable. It too is a specific number. Calculating dz/de makes as much sense as does d2/d3 (which is as meaningless as it looks).

 

[Ambidext]

You should not be looking for f(e). You want some value x such that f(x)=e. In short, you want f^-1(e).

So which means I should equate ln x + e^x = e, and find for x?

 

[D H]

Exactly.

 

[Ambidext]

But that'll give me

e = ln x + e^x
1 = x + ln (ln x)
x =1 - ln (ln x) ?

Sorry my maths is just hopeless.

 

[rock.freak667]

But that'll give me

e = ln x + e^x

Hint: ln(1) = 0, look at what 'x' was replaced with.

 

[Ambidext]

Okay, so

e = e^x + ln x
x = 1,

then how do I find (f^-1)'(e)?

 

[JonF]

use the formula in post #3