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

1 Answer
Aug 25, 2017

Answer:

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)#