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

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Answer 1 · 2016-03-11T00:04:43

$d/dxln((x+1)/(x-1))=1/(x+1)-1/(x-1)$

To avoid needing to use the quotient rule , we will use the property of logarithms that
$log(a/b) = log(a)-log(b)$

After that, we will use the chain rule as well as the known derivative $d/dxln(x) = 1/x$

$d/dxln((x+1)/(x-1)) = d/dx(ln(x+1)-ln(x-1))$

$=d/dxln(x+1)-d/dxln(x-1)$

$=1/(x+1)(d/dx(x+1))-1/(x-1)(d/dx(x-1))$

$=1/(x+1)(1)-1/(x-1)(1)$

$=1/(x+1)-1/(x-1)$

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