What is the derivative of $y = log3 [(x+1/x-1)^ln3]$ ?

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Answer 1 · 2015-09-11T17:18:59

I will guess that this should be: $y = log_3 [(x+1/x-1)^ln3]$ , or $y = log_3 [((x+1)/(x-1))^ln3]$ .

$y = log_3 [(x+1/x-1)^ln3]$

$y =ln3 log_3 (x+1/x-1)$ (power property of logarithms)

$y =ln3 (ln (x+1/x-1))/ln3$ (change of base)

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

So we get:

$y' = 1/(x+1/x-1) * d/dx(x+1/x-1)$

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

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

Note if the original should have been

$y = log_3 [((x+1)/(x-1))^ln3]$ ,

Then we get

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

and

$y' = 1/(x+1)-1/(x-1) = (-2)/(x^2-1)$

What is the derivative of y = log3 [(x+1/x-1)^ln3]y = log3 [(x+1/x-1)^ln3]?