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

Aug 10, 2017

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

#### Explanation:

Since $\frac{1}{x} = {x}^{- 1}$, we can say

$f \left(x\right) = \frac{1}{{x}^{2} + 6} = {\left({x}^{2} + 6\right)}^{- 1}$

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

$\implies f ' \left(x\right) = - 2 x {\left({x}^{2} + 6\right)}^{- 2} = \frac{- 2 x}{{\left({x}^{2} + 6\right)}^{2}}$

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

Another way to write $f ' \left(x\right)$

$f ' \left(x\right) = - 2 x {\left({x}^{2} + 6\right)}^{-} 2$

Then, by the Product Rule $\left(f \left(x\right) g \left(x\right)\right) ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

$\implies f ' ' \left(x\right) = - 2 {\left({x}^{2} + 6\right)}^{-} 2 + 8 {x}^{2} {\left({x}^{2} + 6\right)}^{- 3}$

$= \frac{- 2 \left({x}^{2} + 6\right)}{{\left({x}^{2} + 6\right)}^{3}} + \frac{8 {x}^{2}}{{x}^{2} + 6} ^ 3$

$= \frac{8 {x}^{2} - 2 {x}^{2} - 12}{{\left({x}^{2} + 6\right)}^{3}} = \frac{6 {x}^{2} - 12}{{\left({x}^{2} + 6\right)}^{3}} = \frac{6 \left({x}^{2} - 2\right)}{{\left({x}^{2} + 6\right)}^{3}}$