What is the second derivative of #f(x)=x/(x^2+1)#?

Question

Charles Arocha · Accepted Answer

#f''(x)== (2x^5-4x^3-6x)/(x^2+1)^4#

graph{(2x^5-4x^3-6x)/(x^2+1)^4 [-10, 10, -5, 5]}

#f(x)=x/(x^2+1)# graph{x/(x^2+1) [-10, 10, -5, 5]} #color(red)(f(x)=g(x)/(h(x)##color(red)(=>f'(x)= (g'(x).h(x)-h'(x).g(x))/(f(x))^2# #color(red)(g(x)=x# #color(red)(h(x)=x^2+1# #f'(x)=(1.(x^2+1)-2x(x))/(x^2+1)^2# #=(x^2+1-2x^2)/(x^2+1)^2=(x^2-2x^2+1)/(x^2+1)^2## =(-x^2+1)/(x^2+1)^2# #=># #f'(x)= (-x^2+1)/(x^2+1)^2# #f''(x)= ((-2x).(x^2+1)^2-2(2x)(x^2+1)(-x^2+1))/(x^2+1)^4##;# #color(red)(A(x)=(x^n+r)^m=>A'(x)=m(nx^(n-1))(x^n+r)^(m-1)# #f''(x)= ((-2x).(x^4+2x^2+1)-4xcolor(red)((1-x^4)))/(x^2+1)^4##;# #color(red)((a^2+1)(-a^2+1)=1-a^4# #f''(x)= (-2x^5-4x^3-2x-4x+4x^5)/(x^2+1)^4# #= (color(green)(-2x^5)-4x^3-color(blue)(2x)-color(blue)(4x)+color(green)(4x^5))/(x^2+1)^4# #= (color(red)(2x^5)-4x^3-color(blue)(6x))/(x^2+1)^4# #f''(x)== (color(green)(2x^5)-4x^3-color(blue)(6x))/(x^2+1)^4#

Emily Ammerman · Answer

undefined To find the second derivative of $ f(x) = \frac{x}{x^2 + 1} $, first find the first derivative using the quotient rule:

$ f'(x) = \frac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} $

Simplify this to get:

$ f'(x) = \frac{1 - x^2}{(x^2 + 1)^2} $

Now, differentiate $ f'(x) $ with respect to $ x $ again to find the second derivative:

$ f''(x) = \frac{(1 - x^2)'(x^2 + 1)^2 - (1 - x^2)(2(x^2 + 1)(2x))}{(x^2 + 1)^4} $

Simplify this expression to get the second derivative.