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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.

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up vote 4 down vote accepted

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

$$\eqalign{ \frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + {x^2\over\sqrt{x^2-1}}\Bigr) &= \frac{1}{x \color{maroon}{\sqrt{x^2-1}}} \cdot \color{maroon}{\sqrt{x^2-1} }+ \frac{1}{\color{darkblue}x\color{darkgreen}{ \sqrt{x^2-1}}} \cdot {\color{darkblue}{x^2}\over\color{darkgreen}{\sqrt{x^2-1}}} \cr &={1\cdot\color{maroon}1\over x} +\frac{\color{darkblue}x}{ \color{darkgreen}{{x^2-1}}} \cr &={{(x^2-1)}\cdot1+x\cdot x\over x(x^2-1)}\cr &={2x^2-1\over x(x^2-1)}.\cr } $$

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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.

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I do not know log properties nor do I have time to learn them right now. – user138246 May 10 '12 at 16:05
That's odd since the first thing anyone learns about logarithms is that $\log(ab)=\log(a)+\log(b)$ and $\log(a^c)=c\log(a)$...... – N. S. May 10 '12 at 16:06
I learned that over two years ago and keep forgetting it. Also I did not forget a bracket in my computation, it appears to be correct. – user138246 May 10 '12 at 16:11
It may appear to be correct to you, but it is not. As N.S. says, you have forgotten a bracket. – Chris Taylor May 10 '12 at 16:21
To keep it simple, $[\ln(g)]' = \frac{g'}{g}$. Calculate $g'$ separately and divide it by $g$. Note that $g'$ is calculated with the product rule, as you did, and then EVERYTHING is divided by g. – N. S. May 10 '12 at 16:26

$\log (x\sqrt{x^{2}-1})=\log x+\log(\sqrt{x^{2}-1})=\log x+ \frac{1}{2}\log({x^{2}-1})$ 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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