Thread: Derivatives of natual logs

1. Derivatives of natual logs

$\displaystyle y= \ln (x \sqrt {x^2-1})$

2. Originally Posted by yeloc
$\displaystyle y= \ln (x \sqrt {x^2-1})$
$\displaystyle \frac{d}{dx} \ln(f(x)) = \frac{1}{f(x)} \times \frac{d}{dx} (f(x))$

Hence:

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

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

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

Now use the chain rule again:

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

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

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

3. Thanks! I found my mistake; I wasn't multiplying by the inside function when doing the chain rule.

4. $\displaystyle f(x)=\ln x+\frac{1}{2}\ln \left( {{x}^{2}}-1 \right)\implies f'(x)=\frac{1}{x}+\frac{x}{{{x}^{2}}-1},$ and we're done.