How do you differentiate $f(x)=ln(x/(x-1))$ ?

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Answer 1 · 2015-12-04T10:50:28

$f'(x) = 1/x - 1/(x-1)$

The easiest way is to use the following logarithmic rule first:

$ln(a/b) = ln(a) - ln(b)$

In your case, it means that

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

Now, with the knowledge that $(ln x)' = 1/x$ , you can differentiate your function as follows:

$f'(x) = 1/x - 1/(x-1)$

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