1. ln differentiation

find h'(x) if h(x) = $\ln(x+\sqrt{x^2-1})$
$
h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2})(2x)
$

then I tried to simplify:
$h'(x) = \frac{(2x+x(x^2-1)^\frac{-1}{2})}{x+\sqrt{x^2-1}}$

Is this correct?

2. Originally Posted by littlejodo
find h'(x) if h(x) = $\ln(x+\sqrt{x^2-1})$
$
h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2})(2x)
$

then I tried to simplify:
$h'(x) = \frac{(2x+x(x^2-1)^\frac{-1}{2})}{x+\sqrt{x^2-1}}$

Is this correct?
no, you misplaced the 2x.

it should be: $
h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2}{\color{red}(2x)})
$

now continue

3. Ok, so..

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

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

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

Is this correct? If so, is it simplified enough? Thanks.

4. Originally Posted by littlejodo
Ok, so..

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

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

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

Is this correct? If so, is it simplified enough? Thanks.
that is correct. to me it is simplified enough, i don't know how your professor might like it. s/he might have a problem with that -1/2 power. if you are so inclined, you can multiply the answer by $\frac {(x^2 - 1)^{1/2}}{(x^2 - 1)^{1/2}}$ to get rid of it. but i'd leave it like that