Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 12

In order to find h'(x) we need to know a few things:
The derivative of log(f(x)), and of sqrt(x^2-1).
We need to know the Chain rule. Given h(x)=log(x+sqrt(x^2-1)), we start by differentiating log(f(x)), where f(x) = x+sqrt(x^2-1):
d/dx (log(f(x))) = 1/f(x) * d/dx(f(x)).
Now find the derivative of f(x), that is
d/dx (x+sqrt(g(x))), where g(x) = x^2-1
We have d/dx(x+sqrt(g(x))) = 1 + 1/(2*sqrt(g(x)) * d/dx(g(x))
Finally we are left with finding the derivative of g(x), simple enough.
d/dx (x^2-1) = 2*x
Back substituting we have
d/dx (h(x)) = 1/(x+sqrt(x^2-1)) * (1 + 1/(2*sqrt(x^2-1))*2*x)
Simplifying, h'(x) = 1/(x+sqrt(x^2-1)) * (1 + x/(sqrt(x^2-1))