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.
 A: 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.
A: You obtained the correct derivative, but you need parentheses as such: 
$$\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x \frac{1}{2}(x^2-1)^\frac{-1}{2}2x\Bigr)$$
Clean this up a bit to get
$$\tag{1}
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x^2 (x^2-1)^\frac{-1}{2} \Bigr).
$$
You were ok up to this point. The rest of your work contains an algebraic error: in the second line of the displayed equations after you say "which I reduce to", the second term is off, it needs an "$x$" downstairs.
But other than that, you did fine. For what it's worth here is the derivation with the correction:
Equation ${1}$ can be written as
$$
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + {x^2\over\sqrt{x^2-1}}\Bigr)
$$
Now multiply through
$$\eqalign{
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} +  {x^2\over\sqrt{x^2-1}}\Bigr)
&=
\frac{1}{x \color{maroon}{\sqrt{x^2-1}}} \cdot 
\color{maroon}{\sqrt{x^2-1} }+
\frac{1}{\color{darkblue}x\color{darkgreen}{ \sqrt{x^2-1}}} \cdot {\color{darkblue}{x^2}\over\color{darkgreen}{\sqrt{x^2-1}}} \cr
&={1\cdot\color{maroon}1\over x} +\frac{\color{darkblue}x}{ \color{darkgreen}{{x^2-1}}}  \cr
&={{(x^2-1)}\cdot1+x\cdot x\over x(x^2-1)}\cr
&={2x^2-1\over x(x^2-1)}.\cr
}
$$
A: $\log (x\sqrt{x^{2}-1})=\log x+\log(\sqrt{x^{2}-1})=\log x+ \frac{1}{2}\log({x^{2}-1})$
Let $y=\log x+ \frac{1}{2}\log({x^{2}-1}).$
Now differentiate with respect to $x$, we get,
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{2}\frac{1}{x^{2}-1}2x$
$\therefore \frac{dy}{dx}=\frac{1}{x}+\frac{x}{x^{2}-1}$=$\frac{x^{2}-1+x^{2}}{x(x^{2}-1)}$=$\frac{2x^{2}-1}{x(x^{2}-1)}.$ 
