f(x)=x^(lnx)

This is what I did:

lny = lnx(lnx)

(1/y)y' = (lnx)/x + (lnx)/x

y' = y [ (lnx)/x + (lnx)/x ]

y' = x^(lnx) [ (2lnx)/x)

f'(e) = e^(lne) [(2lne/e)

f'(e) = e^1 (2/e)

f'(e) = 2

Is my method of differentiation correct?

Results 1 to 4 of 4

- August 2nd 2011, 10:38 PM #1

- Joined
- Apr 2011
- Posts
- 58

- August 2nd 2011, 10:58 PM #2

- August 3rd 2011, 07:02 AM #3

- Joined
- Apr 2005
- Posts
- 15,977
- Thanks
- 1643

## Re: f(x)=x^(lnx) , f'(e)=?

- August 3rd 2011, 08:19 AM #4