How do you differentiate $(x(lnx)^(x-1))(1/x)$ ?

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Answer 1 · 2015-04-08T16:28:52

I would first rewrite the function:

$(x(lnx)^(x-1))(1/x) = (lnx)^(x-1)$

Now use logarithmic differentiation:

$y = (lnx)^(x-1)$ , so

$ln[y] = ln[ (lnx)^(x-1)]$ , so

$lny = (x-1) ln(lnx)$ .

Now differentiate implicitly (as we do for logarithmic differentiation).
I use the product rule in the form: $d/(dx)(fg)=f'g+fg'$

Since, $g(x)=ln(lnx)$ we have $g'(x)= 1/ lnx *1/x$

$1/y y' = (1) ln(lnx)+(x-1)[1/ lnx * 1/x]$

$y' = y[ln(lnx)+(x-1)/(x lnx)] = (lnx)^(x-1) [ln(lnx)+(x-1)/(x lnx)]$

How do you differentiate (x(lnx)^(x-1))(1/x)(x(lnx)^(x-1))(1/x)?