Question

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

Answer 1

`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]}

Explanation:

`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`