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

1 Answer
Guilherme N.
Dec 25, 2015

We'll need the chain rule here, which states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

Explanation:

In this case, we must rename #u=(1/x)# and now derivate the function #y=lnu#. Let's do it separately, step-by-step:

#(dy)/(du)=1/u#

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

#(dy)/(dx)=-1/(ux^2)#

Substituting #u#:

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

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