How do you differentiate $y = log_3 [((x+1)/(x-1))^ln3]$ ?

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Answer 1 · 2016-08-28T12:07:16

$2/(1-x^2)$

Use $log_ba=ln a/ln b, ln c^n = n ln c and ln(m/s)=ln m - ln n$

Here,

$y=log_3[((x+1)/(x-1))^ln 3]$

$=ln((x+1)/(x-1))^ln 3/ln 3$

$=ln 3 ln[(x+1)/(x-1)]/ln 3$

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

y'=1/(x+1)(x+1)'-1/(x-1)(x-1)'#

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

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

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

How do you differentiate y = log_3 [((x+1)/(x-1))^ln3]y = log_3 [((x+1)/(x-1))^ln3]?