If u = log(sqrt{x-1} - sqrt{x+1}) and v = sqr | vedclass

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(D) Given $u = \log(\sqrt{x-1} - \sqrt{x+1})$ and $v = \sqrt{x+1} + \sqrt{x-1}$.Note that $(\sqrt{x+1} + \sqrt{x-1})(\sqrt{x+1} - \sqrt{x-1}) = (x+1)

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$\frac{-1}{v}$

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(D) Given $u = \log(\sqrt{x-1} - \sqrt{x+1})$ and $v = \sqrt{x+1} + \sqrt{x-1}$.Note that $(\sqrt{x+1} + \sqrt{x-1})(\sqrt{x+1} - \sqrt{x-1}) = (x+1) - (x-1) = 2$.Thus,$\sqrt{x+1} - \sqrt{x-1} = \frac{2}{v}$.However,the argument of the log is $\sqrt{x-1} - \sqrt{x+1} = -(\sqrt{x+1} - \sqrt{x-1}) = -\frac{2}{v}$.Since the domain of $\log$ requires a positive argument,we consider the magnitude or the complex form. Assuming the standard derivative approach:$u = \log(-2) - \log(v)$.Taking the derivative with respect to $v$:$\frac{du}{dv} = \frac{d}{dv}(\log(-2) - \log(v)) = 0 - \frac{1}{v} = -\frac{1}{v}$.

If $u = \log(\sqrt{x-1} - \sqrt{x+1})$ and $v = \sqrt{x+1} + \sqrt{x-1}$,then $\frac{du}{dv} = \dots$.

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