How do you find the first and second derivative of #ln(x/(x^2+1))#?

1 Answer
Nov 21, 2016

Answer:

#f(x)=ln(frac{x}{x^2+1})#

#f'(x)=[-x^2+1]/[x^3+x]#

#f''(x)=frac{x^4-2x^2-1}{(x^3+x)^2}#

Explanation:

Let #f(x)=ln(frac{x}{x^2+1})#

#f'(x)=(frac{(x^2+1)(1)-(x)(2x)}{(x^2+1)^2})((x^2+1)/x)#
#f'(x)=(frac{x^2+1-2x^2}{(x^2+1)^2})((x^2+1)/x)#
#f'(x)=(frac{-x^2+1}{(x^2+1)^2})((x^2+1)/x)#
#f'(x)=[-x^2+1]/[x(x^2+1)]#

#f'(x)=[-x^2+1]/[x^3+x]#

#f''(x)=frac{(x^3+x)(-2x)-(-x^2+1)(3x^2+1)}{(x^3+x)^2}#
#f''(x)=frac{(-2x^4-2x^2)-(-3x^4-x^2+3x^2+1)}{(x^3+x)^2}#
#f''(x)=frac{-2x^4-2x^2+3x^4-2x^2-1}{(x^3+x)^2}#

#f''(x)=frac{x^4-2x^2-1}{(x^3+x)^2}#