What is the derivative of #ln(1/x)#?

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Answer 1 · 2018-04-13T00:56:00

#d/dxln(1/x)=-1/x#

We could use the Chain Rule right away, but the properties of logarithms allow us to avoid that and make this easier.

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

So,

#d/dxln(1/x)=d/dx(-lnx)=-1/x#

Answer 2 · 2018-04-13T04:29:09

#-1/x#

Well, let's try the chain rule, which states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=1/x,:.(du)/dx=-1/x^2#.

Then #y=lnu,:.dy/(du)=1/u#.

Combining, we get,

#dy/dx=1/u*-1/x^2#

#=-1/(ux^2)#

Substituting back #u=1/x#, we get,

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

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

#=-1/x#