Question

What are the first and second derivatives of `f(x)=1/(x^2+6)`?

Answer 1

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

Explanation:

Since `1/x=x^(-1)`, we can say

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

Then, by the chain rule `(f(g(x))'=f'(g(x))g'(x)`

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

To get the 2nd derivative, we simply take the derivative of the derivative

Another way to write `f'(x)`

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

Then, by the Product Rule `(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)`

`=> f''(x)=-2(x^2+6)^-2+8x^2(x^2+6)^(-3)`

`=(-2(x^2+6))/((x^2+6)^3)+(8x^2)/(x^2+6)^3`

`=(8x^2-2x^2-12)/((x^2+6)^3)=(6x^2-12)/((x^2+6)^3)=(6(x^2-2))/((x^2+6)^3)`