Question

How do you find the second derivative of `y=ln((2x)/(x+3))` ?

Answer 1

`1/(x+3)^2-1/x^2`

Explanation:

`y= ln 2 + ln x - ln (x|3)`

`y'=0+1/x-1/(x+3)=x^(-1)-(x+2)^(-1)`

`y''=(-1)x^(-1-1)-(-1)(x+2)^(-1-1)(x+3)'`

`=-1/x^2+1/(x+3)^2(1)`