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

Question

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

I was trying to do the differentiation for HW. I managed to get $$\frac{ 2 \sqrt{x^2 - 1} - x}{ 2 \sqrt{x^2 - 1} \, (x - \sqrt{x^2 - 1} ) }.$$ But the answer is supposed to be $$\frac{-1}{\sqrt{x^2 - 1}}.$$

Need a hint on what to do next.

Answer 1

applying $\ln(u)'=\frac{u'}{u}$ you should get something not quite like you got (not going to do the work for you…) and then factorize $\sqrt {{x}^{2}-1}$ at denominator.

Answer 2

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

Is a fraction of

$$ 1- \frac {x}{\sqrt {x^2-1}}$$

Over $${x-\sqrt {x^2-1}}$$

Upon simplification you get the answer

$$\frac {-1}{\sqrt{x^2-1}}$$