How do you differentiate #(lnx)^(x)#?

1 Answer
Steve M
Dec 24, 2016

# d/dx(lnx)^x = (lnx)^x{1/lnx + ln((lnx))} #

Explanation:

Let #y=(lnx)^x#

Take (Natural) logarithms of both sided:

# " " lny = ln((lnx)^x ) #
# :. lny = xln((lnx) ) #

Differentiate Implicitly (LHS) and apply product rule and chain rule (RHS).

# \ \ \ \ \ \ 1/ydy/dx = (x)(1/lnx*1/x) + (1)(ln((lnx)) #
# :. \ 1/ydy/dx = 1/lnx + ln((lnx)) #
# :. " " dy/dx = y{1/lnx + ln((lnx))} #
# :. " " dy/dx = (lnx)^x{1/lnx + ln((lnx))} #

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