Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

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

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

share|improve this answer
add comment

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.

share|improve this answer
    
I do not know log properties nor do I have time to learn them right now. –  user138246 May 10 '12 at 16:05
2  
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
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.