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

1 Answer
Mar 18, 2017

Answer:

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

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

Explanation:

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)#