<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] $\log (x\sqrt{x^{2}-1})=\log x+\log(\sqrt{x^{2}-1})=\log x+ \frac{1}{2}\log({x^{2}-1})$
Let $y=\log x+ \frac{1}{2}\log({x^{2}-1}).$
Now differentiate with respect to $x$, we get,
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{2}\frac{1}{x^{2}-1}2x$
$\therefore \frac{dy}{dx}=\frac{1}{x}+\frac{x}{x^{2}-1}$=$\frac{x^{2}-1+x^{2}}{x(x^{2}-1)}$=$\frac{2x^{2}-1}{x(x^{2}-1)}.$ 
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