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

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Answer 1 · 2017-03-18T15:52:10

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

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

Since
$f(x)=ln(g(x))->f'(x)=1/g(x)*g'(x)$

and

$g(x)=(n(x))/(d(x))->g'(x)=(n'(x)d(x)-n(x)d'(x))/(d(x))^2$ ,

the first derivative is:

$f'(x)=cancel((x^2-1))/x*(1(x^2-1)-x(2x))/(x^2-1)^cancel2$

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

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

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

$f'(x)=(1+x^2)/(x-x^3)$

The second derivative is:

$f''(x)=(2x^2(1-x^2)-(1+x^2)(1-3x^2))/(x^2(1-x^2)^2)$

$=(2x^2-2x^4-1-x^2+3x^2+3x^4)/(x^2(1-x^2)^2$

$=(x^4+4x^2-1)/(x^2(1-x^2)^2)$

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