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)]#