# How do you find the first and second derivative of #y=ln [ x / (x^2 - 1) ]#?

and

the first derivative is:

The second derivative is:

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To find the first and second derivative of ( y = \ln\left(\frac{x}{x^2 - 1}\right) ):

First, find the first derivative ( y' ): [ y' = \frac{d}{dx} \ln\left(\frac{x}{x^2 - 1}\right) ]

Apply the chain rule and the derivative of ln: [ y' = \frac{1}{\frac{x}{x^2 - 1}} \cdot \frac{d}{dx} \left(\frac{x}{x^2 - 1}\right) ] [ y' = \frac{x^2 - 1}{x} \cdot \frac{d}{dx} \left(\frac{x}{x^2 - 1}\right) ] [ y' = \frac{x^2 - 1}{x} \cdot \left(\frac{(x^2 - 1) - x(2x)}{(x^2 - 1)^2}\right) ] [ y' = \frac{x^2 - 1}{x} \cdot \left(\frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2}\right) ] [ y' = \frac{x^2 - 1}{x} \cdot \left(\frac{-x^2 - 1}{(x^2 - 1)^2}\right) ] [ y' = -\frac{x^2 - 1}{x(x^2 - 1)} ]

Now, find the second derivative ( y'' ): [ y'' = \frac{d}{dx} \left(-\frac{x^2 - 1}{x(x^2 - 1)}\right) ] [ y'' = \frac{(-x^2 - 1)(x(x^2 - 1)) - (-x^2 - 1)(x^2 - 1)}{(x(x^2 - 1))^2} ] [ y'' = \frac{-(x^3 - x + x^3 - x)}{(x^2(x^2 - 1))^2} ] [ y'' = \frac{-2x^3 + 2x}{x^4(x^2 - 1)} ] [ y'' = \frac{-2x(x^2 - 1)}{x^4(x^2 - 1)} ] [ y'' = -\frac{2}{x^3} ]

So, the first derivative is ( y' = -\frac{x^2 - 1}{x(x^2 - 1)} ) and the second derivative is ( y'' = -\frac{2}{x^3} ).

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