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

1 Answer
Jul 28, 2014

#y'=-1/x#

Full solution

#y=ln(1/x)#

This can be solved in two different ways,

Explanation (I)

The simplest one is, using logarithm identity,

#log(1/x^y)=log(x^-y)=-ylog (x)#, similarly following for problem,

#y=-ln(x)#

#y'=-(ln(x))'#

#y'=-1/x#

Explanation (II)

Using Chain Rule,

#y'=(ln(1/x))'#

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

#y'=-1/x#