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

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

- Joined
- Apr 2011
- Posts
- 58

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

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

- Joined
- Apr 2005
- Posts
- 19,767
- Thanks
- 3027

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

At this point it would be slightly simpler to write $\displaystyle ln(y)= (ln(x))^2$ so that

$\displaystyle \frac{1}{y}y'= \frac{2ln(x)}{x}$ by the chain rule.

(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?

- Aug 3rd 2011, 08:19 AM #4
## Re: f(x)=x^(lnx) , f'(e)=?