Question

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

Answer 1

`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`