Question

Given that `y=ln((x-3)/(x+3))`, how do you show that `(dy)/(dx)=6/(x^2-9)`?

Answer 1

See below

Explanation:

`y=ln((x-3)/(x+3))`

`=ln(x-3) - ln (x+3)`

`implies y' = 1/(x-3) - 1/(x+3)`

`= (x + 3 - (x-3))/((x-3)(x+3))`

`= 6/((x-3)(x+3))`

QED