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 $d/dx(-ln(x))=-1/x$ .

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