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

Aug 25, 2017

Recall that $\frac{d}{\mathrm{du}} \ln \left(u\right) = \frac{\mathrm{du}}{u}$ 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 $\ln \left(u\right)$ is $\frac{\mathrm{du}}{u}$. 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)