Derivatives of natual logs

• Feb 5th 2009, 01:37 PM
yeloc
Derivatives of natual logs
$y= \ln (x \sqrt {x^2-1})$
• Feb 5th 2009, 01:45 PM
Mush
Quote:

Originally Posted by yeloc
$y= \ln (x \sqrt {x^2-1})$

$\frac{d}{dx} \ln(f(x)) = \frac{1}{f(x)} \times \frac{d}{dx} (f(x))$

Hence:

$\frac{d}{dx} \ln(x\sqrt{x^2-1}) = \frac{1}{x\sqrt{x^2-1}} \times \frac{d}{dx} (x\sqrt{x^2-1})$

$= \frac{1}{x\sqrt{x^2-1}} \times \frac{d}{dx} (\sqrt{x^4-x^2})$

$= \frac{1}{x\sqrt{x^2-1}} \times \frac{d}{dx} ({x^4-x^2})^{\frac{1}{2}}$

Now use the chain rule again:

$= \frac{1}{x\sqrt{x^2-1}} \times \frac{1}{2}(x^4-x^2)^{\frac{-1}{2}} \times \frac{d}{dx} (x^4-x^2)$

$= \frac{1}{x\sqrt{x^2-1}} \times \frac{1}{2\sqrt{x^4-x^2}} \times \frac{d}{dx} (x^4-x^2)$

$= \frac{1}{x\sqrt{x^2-1}} \times \frac{1}{2x\sqrt{x^2-1}} \times \frac{d}{dx} (x^4-x^2)$
• Feb 5th 2009, 02:39 PM
yeloc
Thanks! I found my mistake; I wasn't multiplying by the inside function when doing the chain rule.
• Feb 5th 2009, 03:29 PM
Krizalid
$f(x)=\ln x+\frac{1}{2}\ln \left( {{x}^{2}}-1 \right)\implies f'(x)=\frac{1}{x}+\frac{x}{{{x}^{2}}-1},$ and we're done.