y =ln((x^2+1)/(x^2-1))^1/2

Question

Find the first derivative for this ln square division.

Answer 1

y =ln((x^2+1)/(x^2-1))^1/2
y = 1/2 ln((x^2+1)/(x^2-1))
y = 1/2(ln(x^2+1) - ln(x^2-1))

y' = 1/2 (2x/(x^2+1) - 2x/(x^2-1))
y' = x/(x^2+1) - x/(x^2-1)
y' = -2x/(x^4-1)

Now, if you're a glutton for punishment, just apply the chain rule:

y =ln((x^2+1)/(x^2-1))^1/2
y' = (1/2)/[(x^2+1)/(x^2-1)]^(-1/2) * [(2x)(x^2-1) - 2x(x^2+1)]/(x^2-1)^2

I'm not!

Answer 2

Hmm. I seem to have misread the problem. You have the square root of the log?

In that case,

y' = -2x/[(x^4-1)*ln((x^2+1)/(x^2-1))^1/2]

go to wolframalpha and show steps