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.

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- Oct 31st 2012, 09:17 AMaimfortheheadLogarithmic 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 AMSironRe: 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 AMaimfortheheadRe: 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 PMskeeterRe: Logarithmic differentiation
$\displaystyle y = (\ln{x})^x$

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

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

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

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

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

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