What is the derivative of #ln(lnx^2)#?

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Answer 1 · 2016-03-15T06:22:27

#d/dx [ln(lnx^2)]# = #1/[xln(x)]#

#y = ln(lnx^2)#

Using the chain rule:

#d/dx[ln(f(x)]# = #[lnf(x)]^' = [1/f(x)]f'(x)#

#dy/dx = [1/(lnx^2)] (lnx^2)^'#

#= [1/(lnx^2)] (1/x^2)(x^2)^'#

#=[1/(lnx^2)] (1/x^2)(2x)#

#=2/[xln(x^2)]# => simplifying:

#=1/[xln(x)]#