Question

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

Answer 1

`(d^2y)/dx^2=(2-2x^2)/(x^2+1)^2`

Explanation:

let `y=ln(x^2+1)`
this is a chain differentiation of `ln(u(x))`
`dy/dx=(u'(x))/(u(x))`
So `dy/dx=(2x)/(x^2+1)`
tHen we use `(u/v)'=(u'v-uv')/v^2`
`(d^2y)/dx^2=(2(x^2+1)-2x*2x)/(x^2+1)^2=(2x^2+2-4x^2)/(x^2+1)^2=(2-2x^2)/(x^2+1)^2`