What is the derivative of #lnx /x#?

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Answer 1 · 2018-04-08T23:56:10

#d/dxlnx/x=1/x^2(1-lnx)#

We could use the Quotient Rule, but it's nice to avoid it where it is unnecessary. Instead, we can rewrite as

#x^-1lnx# and differentiate with the Product Rule, recalling that #d/dxlnx=1/x=x^-1#

#d/dxlnx/x=x^-1(d/dxlnx)+lnx(d/dxx^-1)#

#=x^-1x^-1-x^-2lnx#

#x^-1x^-1=x^(-1-1)=x^-2=1/x^2#

Thus,

#d/dxlnx/x=1/x^2-lnx/x^2#

#d/dxlnx/x=1/x^2(1-lnx)#