Question

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

Answer 1

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

Explanation:

`f(x)=ln(x^2+2)`

Using the chain rule we have:
`f'(x)=1/(x^2+2) *(2x)`
`:. f'(x)=(2x)/(x^2+2)`

To find the second derivative we now need to use the product rule. Remember:
`d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2`; so

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