How to differentiate ln $\frac{x}{2}$ $x/2$ and ln( $\sqrt (1 -$ $√(1-$ 1/x $)$ $)$ ) respect to x?

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How to differentiate ln $\frac{x}{2}$ $x/2$ and ln( $\sqrt (1 -$ $√(1-$ 1/x $)$ $)$ ) respect to x?

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Answer 1 · 2018-05-13T16:08:51

a) Simplify using logarithm rules:

$ln (\frac{x}{2}) = ln (x) - ln (2)$ $ln(x/2) = ln(x) - ln(2)$

$\frac{d}{d x} (ln (\frac{x}{2})) = \frac{1}{x} - 0 = \frac{1}{x}$ $d/dx(ln (x/2)) = 1/x - 0 = 1/x$

b) Once again apply logarithm laws to your advantage

$ln \sqrt{(1 - \frac{1}{x})} = \frac{1}{2} ln (1 - \frac{1}{x})$ $lnsqrt((1 - 1/x)) = 1/2ln(1 - 1/x)$

Now by the chain rule:

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

Hopefully this helps!

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