Question

How do you find the first and second derivative of `ln(x+sqrt((x^2)-1))`?

Answer 1

Recall that `d/(du) ln (u) =(du)/u` first. We use this and the chain rule, giving us a first derivative of `(1+x (x^2-1)^-(1/2))/(x+(x^2-1)^(1/2)`

Explanation:

The derivative of `ln (u)` is `(du)/u`. In this case, `u=(x+sqrt ((x^2)-1) = x +((x^2)-1)^(1/2)`

The derivative of this expression will require use of the chain rule on our square root...

`(du)/(dx) = d/(dx) (x + ((x^2)-1)^(1/2) = 1 +x (x^2 -1)^(-1/2)`

This is our du term. Our u is already given, so du/u is

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