Question

How do you find the first and second derivative of `lnx^2`?

Answer 1

`y'=2x/x^2`

`y''=2/x^2`

Explanation:

Using the chain rule we get:

`y'=(1/x^2)*2x`=`2x/x^2`

`y''=((x^2*2)-4x^2)/x^4`=`(2x^2-4x^2)/x^4`=`-2x^2/x^4`=`2/x^2`