How do you find the first and second derivative of y=ln [ x / (x^2 - 1) ]y=ln[xx21]?

1 Answer
Mar 18, 2017

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)