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)#