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

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Answer 1 · 2017-08-25T18:56:31

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

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

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

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