Logarithmic differentiation

• Oct 31st 2012, 09:17 AM
Logarithmic differentiation
Can someone tell me if this is correct?
y = (lnx)^x
y' = x(lnx)^x-1 * 1/x
y' = (lnx) ^ x-1

It seemed a bit too simple so I wanted to check if I'm doing it right, thanks.
• Oct 31st 2012, 09:35 AM
Siron
Re: Logarithmic differentiation
It's not correct.
Write $\displaystyle y = [\ln(x)]^x = e^{x\ln[\ln(x)]}$
Now, compute $\displaystyle \frac{dy}{dx}$
• Oct 31st 2012, 09:39 AM
Re: Logarithmic differentiation
I'm trying to understand what you did there, but I also have a second method,
Which is: lny = xln(lnx)
Would this also be correct?
So...
y'/y = ln(lnx) + x * 1/lnx * 1/x
y' = y(ln(lnx) + 1/lnx)
y' = (lnx)^x(ln(lnx) + 1/lnx)
Or is this also wrong?
• Oct 31st 2012, 04:19 PM
skeeter
Re: Logarithmic differentiation
$\displaystyle y = (\ln{x})^x$

$\displaystyle \ln{y} = \ln(\ln{x})^x$

$\displaystyle \ln{y} = x \cdot \ln(\ln{x})$

$\displaystyle \frac{y'}{y} = x \cdot \frac{1}{x\ln{x}} + \ln(\ln{x}) \cdot 1$

$\displaystyle \frac{y'}{y} = \frac{1}{\ln{x}} + \ln(\ln{x})$

$\displaystyle y' = y\left[\frac{1}{\ln{x}} + \ln(\ln{x})\right]$

$\displaystyle y' = (\ln{x})^x\left[\frac{1}{\ln{x}} + \ln(\ln{x})\right]$