Question

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

Answer 1

`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}`