How do you differentiate $y = ln (\frac{1}{x})$ $y= ln( 1/x)$ ?

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Answer 1 · 2016-07-31T20:44:40

$= - \frac{1}{x}$ $= - 1/x$

doing it the long way:

$d/dx ln (f(x)) = 1/(f(x)) * f'(x)$

here $f(x) = 1/x = x^(-1)$ so by the power rule $d/dx (x^(-1)) = - x^(-2) = -1/x^2$

so

$d/dx ln (1/x) = 1/(1/x) * -1/x^2 = - 1/x$

speeding up a little:

$ln (1/x) = ln x^(-1) = - ln x$

$d/dx (- ln x) = - d/dx (ln x) = - 1/x (x)' = - 1/x$

How do you differentiate y= ln( 1/x)y=ln(1x)y= ln( 1/x)?