What is the derivative of $log_2(x^2/(x-1))$ ?

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Answer 1 · 2016-11-21T03:18:32

$2/(xln(2))- 1/((x - 1)ln(2))$

$log_2(x^2/(x - 1)) =$

$log_2(x^2)- log_2(x - 1) =$

$2log_2(x)- log_2(x - 1) =$

$(2/ln(2))ln(x)- 1/ln(2)ln(x - 1)$

Differentiate:

$(2/ln(2))1/(x)- 1/ln(2)1/(x - 1) =$

Simplify:

$2/(xln(2))- 1/((x - 1)ln(2))$

Answer 2 · 2016-11-21T03:37:48

$=1/ln2((x-2)/(x(x-1)))$

Use $log_ba=lna/lnb$ .

$(log_2(x^2/(x-1))'$

$=(ln (x^2/(x-1))/(ln2))'$

$=1/ln2(lnx^2-ln(x-1))'$

$1/ln2(2(lnx)'-(ln(x-1))')$

$=1/ln2(2/x-1/(x-1))$

$=1/ln2((x-2)/(x(x-1)))$

What is the derivative of log_2(x^2/(x-1))log_2(x^2/(x-1))?