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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- October 31st 2012, 10:17 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. - October 31st 2012, 10:35 AMSironRe: Logarithmic differentiation
It's not correct.

Write

Now, compute - October 31st 2012, 10:39 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? - October 31st 2012, 05:19 PMskeeterRe: Logarithmic differentiation