# Math Help - f(x)=x^(lnx) , f'(e)=?

1. ## f(x)=x^(lnx) , f'(e)=?

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?

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

Originally Posted by NeoSonata
Is my method of differentiation correct?
Yes, it is correct.

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

Originally Posted by NeoSonata
f(x)=x^(lnx)

This is what I did:

lny = lnx(lnx)
At this point it would be slightly simpler to write $ln(y)= (ln(x))^2$ so that
$\frac{1}{y}y'= \frac{2ln(x)}{x}$ 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?

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

Originally Posted by HallsofIvy
At this point it would be slightly simpler to write $ln(y)= (ln(x))^2$ so that
$\frac{1}{y}y'= \frac{2ln(x)}{x}$ by the chain rule.
I noticed that, too, but thought that it would be simpler to let someone else type it up!
To move this reply in the direction of legitimacy, let me also comment that

$(ln(x))^2 \; \;$ is sometimes written as $ln^2(x)$