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

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Answer 1 · 2017-12-08T08:22:54

The answer is $= - \frac{1}{\sqrt{x^{2} - 1}}$ $=-1/sqrt(x^2-1)$

Let $f (x) = ln (x - \sqrt{x^{2} - 1})$ $f(x)=ln(x-sqrt(x^2-1))$

The derivative of $ln u (x)$ $lnu(x)$ is $(u'(x))/(u(x))$

$ln(u(x))'=(u'(x))/(u(x))$

$(sqrtv(x))'=(v'(x))/(2(sqrtv(x)))$

Therefore,

$f'(x)=1/((x-sqrt(x^2-1)))*(1-1/(2sqrt(x^2-1))*2x)$

$=1/((x-sqrt(x^2-1)))*(1-x/sqrt(x^2-1))$

$=1/((x-sqrt(x^2-1)))*(sqrt(x^2-1)-x)/sqrt(x^2-1)$

$=-1/sqrt(x^2-1)$