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

Question

Paisley Johnson · Accepted 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}# 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}#

Ezekiel Alexander · Answer

undefined To find the first and second derivatives of ln(x/(x^2+1)), we use the quotient rule for differentiation:

1. First Derivative:
   f'(x) = [1/(x/(x^2+1))][(x/(x^2+1))' - (x^2+1)'(x/(x^2+1))^2]
         = [(x^2+1) - x(2x)] / [(x^2+1)^2]
         = (x^2 + 1 - 2x^2) / (x^2 + 1)^2
         = (1 - x^2) / (x^2 + 1)^2

2. Second Derivative:
   To find the second derivative, differentiate the first derivative with respect to x:
   f''(x) = [(2x)(x^2 + 1)^2 - (1 - x^2)(2(x^2 + 1)(2x))] / (x^2 + 1)^4
          = (2x(x^2 + 1)^2 - 2(1 - x^2)(2x^3 + 2x)) / (x^2 + 1)^4
          = (2x(x^2 + 1)^2 - 4x(1 - x^2)(x^2 + 1)) / (x^2 + 1)^4
          = (2x(x^4 + 2x^2 + 1) - 4x(x^2 + 1 - x^4 - 1)) / (x^2 + 1)^4
          = (2x^5 + 4x^3 + 2x - 4x^3 - 4x) / (x^2 + 1)^4
          = (2x^5 - 4x^4 - 2x) / (x^2 + 1)^4