<s>[INST] Derivative of $\ln(x\sqrt{x^2-1})$ I am trying to find the derivative of $\ln(x\sqrt{x^2-1})$ but I can not get what the book gets.
I get $$\frac{1}{x \sqrt{x^2-1}} \cdot \sqrt{x^2-1} + x\cdot\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x$$
which I reduce to
$$\begin{align}
&\frac{1}{x\sqrt{x^2-1}}\sqrt{x^2-1} + x^2(x^2-1)^\frac{-1}{2}=\\
&\frac{\sqrt{x^2-1}}{x\sqrt{x^2-1}}  + \frac{x^2(x^2-1)^\frac{-1}{2}}{\sqrt{x^2-1}}=\\
&\frac{1}{x} + \frac{x^2}{x^2-1}= \frac{x^2 - 1 +x^3}{x^3 - x}
\end{align}$$
From here I am not sure what to do. This is not the right answer and I do not know what to do.
 [/INST] You forgot a bracket:
$$\frac{1}{x \sqrt{x^2-1}} * \left[ \sqrt{x^2-1} + x*\frac{1}{2}(x^2-1)^\frac{-1}{2}2x \right]$$
Also, might be much easier to use properties of Log:
$$\ln(x\sqrt{x^2-1}) = \ln(x) +\frac{1}{2} \ln(x^2-1) \,$$
This is much easier to differentiate.
</s>