How do you differentiate #ln(1/x)#?

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Answer 1 · 2016-02-28T21:04:53

It is

#ln(1/x)=-lnx#

hence #d(ln(1/x))/dx=d(-lnx)/dx=-(d(lnx))/dx=-1/x#

Note we used the following identity

#ln(a/b)=lna-lnb#

Answer 2 · 2016-02-29T23:00:44

#-1/x#

Instead of breaking #ln(1/x)# into #ln(1)-ln(x)=-ln(x)#, we can also use the rule that

#ln(a^b)=b*ln(a)#,

thus

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

Since #d/dx(ln(x))=1/x#, we see that #d/dx(-ln(x))=-1/x#.