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Math Help - ln differentiation

  1. #1
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    ln differentiation

    find h'(x) if h(x) = \ln(x+\sqrt{x^2-1})
    <br />
h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2})(2x)<br />

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

    Is this correct?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by littlejodo View Post
    find h'(x) if h(x) = \ln(x+\sqrt{x^2-1})
    <br />
h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2})(2x)<br />

    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: <br />
h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2}{\color{red}(2x)})<br />

    now continue
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  3. #3
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    Ok, so..

    h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2}{(2x)})<br />

    h'(x) = \frac{(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2}{(2x)})}{{x+\sqrt{x^2-1}}}<br />

    h'(x) = \frac{(1 + x(x^2-1)^\frac{-1}{2})}{{x+\sqrt{x^2-1}}}<br />

    Is this correct? If so, is it simplified enough? Thanks.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by littlejodo View Post
    Ok, so..

    h'(x) = \frac{1}{x+\sqrt{x^2-1}}(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2}{(2x)})<br />

    h'(x) = \frac{(1 + \frac{1}{2}(x^2-1)^\frac{-1}{2}{(2x)})}{{x+\sqrt{x^2-1}}}<br />

    h'(x) = \frac{(1 + x(x^2-1)^\frac{-1}{2})}{{x+\sqrt{x^2-1}}}<br />

    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
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