Question

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

Answer 1

`f''(x)=(12x(x^2-1))/(x^2+1)^4`

Explanation:

Use the quotient rule: for a function `f(x)=(g(x))/(h(x))`,

`f'(x)=(g'(x)h(x)-h'(x)g(x))/(h(x))^2`

`g(x)=x`
`h(x)=(x^2+1)^2`

`g'(x)=1`
`h'(x)=4x(x^2+1)`

Thus,

`f'(x)=(1(x^2+1)^2-4x(x^2+1)(x))/(x^2+1)^4`

Simplify.

`f'(x)=((x^2+1)(x^2+1-4x^2))/(x^2+1)^4`

`f'(x)=(1-3x^2)/(x^2+1)^3color(white)(sss)` First Derivative

To find the second derivative, use the quotient rule again.

`g(x)=1-3x^2`
`h(x)=(x^2+1)^3`

`g'(x)=-6x`
`h'(x)=6x(x^2+1)^2`

Thus,

`f''(x)=(-6x(x^2+1)^3-6x(x^2+1)^2(1-3x^2))/(x^2+1)^6`

Simplify.

`f''(x)=(-6x(x^2+1)^2(x^2+1+1-3x^2))/(x^2+1)^6`

`f''(x)=(-6x(-2x^2+2))/(x^2+1)^4`

`f''(x)=(12x(x^2-1))/(x^2+1)^4color(white)(sss)` Second Derivative

The main two pitfalls here are

1. algebra
2. remembering to use the chain rule when differentiating things like `(x^2+1)^2`